Daily New User Report | Sandbox

Identifying similar polygons

The diagram below shows two polygons. The figure is drawn to scale.

Determine whether the two shapes shown are similar or not.
Select the correct reason which explains the first answer.


shape A	shape B

Answer:

Are the shapes similar?
The reason is:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Similar objects must have the same shape, but might not be the same size.

STEP: Question 1: Compare the angles and sides of the two shapes
[−1 point ⇒ 1 / 2 points left]

The question shows us two shapes, and we need to decide if they are similar or not. In maths, two shapes are similar if corresponding angles are equal and if corresponding sides are all proportional (have the same ratio). In other words, if two objects are similar to each other, one of them can be "zoomed in or out" to make it identical to the other one.

These shape are parallelograms, and they are similar. We can see that the four angles are all equal, and the sides are all proportional: $\frac{4}{6} = \frac{10}{15}$ .


shape A	shape B

These two shapes are similar. The correct choice from the list is: Yes.

STEP: Question 2: Choose the correct reason or reasons
[−1 point ⇒ 0 / 2 points left]

As noted in the above explanation, the shapes are similar. This can only be true if the corresponding angles are equal and the sides are all proportional: both must be true for similarity.

Therefore, the correct choice from the list is: Both of the above

Submit your answer as: and

Identifying similar polygons

The diagram below shows two polygons. The figure is drawn to scale.

Are the shapes here similar to one another (select from the choices below)?
Select the correct reason which explains the first answer.


shape A	shape B

Answer:

Are the shapes similar?
The reason is:

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Similar objects must have the same shape, but might not be the same size.

STEP: Question 1: Compare the angles and sides of the two shapes
[−1 point ⇒ 1 / 2 points left]

The question shows us two shapes, and we need to decide if they are similar or not. In maths, two shapes are similar if corresponding angles are equal and if corresponding sides are all proportional (have the same ratio). In other words, if two objects are similar to each other, one of them can be "zoomed in or out" to make it identical to the other one.

These shapes are similar: the angles are all 90 degrees, and the sides are proportional because the sides of each shape are all equal: $\frac{3}{6} = \frac{3}{6}$ . In fact, any two equilateral squares must be similar.


shape A	shape B

These two shapes are similar. The correct choice from the list is: Yes.

STEP: Question 2: Choose the correct reason or reasons
[−1 point ⇒ 0 / 2 points left]

As noted in the above explanation, the shapes are similar. This can only be true if the corresponding angles are equal and the sides are all proportional: both must be true for similarity.

Therefore, the correct choice from the list is: Both of the above

Submit your answer as: and

Identifying similar polygons

The diagram below shows two polygons. The figure is drawn to scale.

Are the polygons here similar to one another (select from the choices below)?
What is the ratio of the size of polygon A to polygon B. Give your answer as a fraction. If the polygons are not similar, write 'None' for the ratio.


polygon A	polygon B

Answer:

Are the shapes similar?
The ratio of the lengths is:

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Similar objects must have the same shape, but might not be the same size.

STEP: Question 1: Compare the angles and sides of the two polygons
[−1 point ⇒ 1 / 2 points left]

The question shows us two shapes, and we need to decide if they are similar or not. In maths, two shapes are similar if corresponding angles are equal and if corresponding sides are all proportional (have the same ratio). In other words, if two objects are similar to each other, one of them can be "zoomed in or out" to make it identical to the other one.

These polygons are similar. In fact, they are equal (identical). If two objects are equal then they must be similar because the angles and sides are all equal.


polygon A	polygon B

These two polygons are similar. The correct choice from the list is: Yes.

STEP: Question 2: Calculate the ratio of the shapes (if they are similar)
[−1 point ⇒ 0 / 2 points left]

Similar shapes always have proportional sides. In this case, the polygons have exactly the same shape, so the ratio of the sizes is $1$ .

The correct answer is: $1$ .

Submit your answer as: and

Prove similarity in adjacent triangles

Consider the following diagram:

Prove that the triangles are similar.

Chukuemeka has already answered the question, and his proof is written below. But, Chukuemeka has made a mistake. Look carefully at his proof and identify where he has made his mistake.

Line	Proof
(a)	In $\Delta LJM$ and $\Delta LKJ$ :
(b)	1. $\dfrac{MJ}{JK} = \dfrac{\text{20}}{\text{12}} = \dfrac{\text{5}}{\text{3}}$
(c)	2. $\dfrac{ML}{JL} = \dfrac{\text{25}}{\text{15}} = \dfrac{\text{5}}{\text{3}}$
(d)	3. $\dfrac{LK}{LJ} = \dfrac{\text{15}}{\text{9}} = \dfrac{\text{5}}{\text{3}}$
(e)	$\therefore \Delta LJM \;\|\|\|\; \Delta LKJ$ (sides of $\Delta$ in prop)

Answer:

The mistake is on Line .

Replace this line with

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Check each line of the proof carefully. Are the sides of the triangles being divided in a consistent order? Is the similarity stated correctly, with the correct reason?

STEP: Identify the error in the proof
[−1 point ⇒ 2 / 3 points left]

When we use sides of $\Delta$ in prop to prove that two triangles are similar, we must divide the sides in the same way each time.

Here, Chukuemeka divided the sides in $\Delta LJM$ by the sides in $\Delta LKJ$ on Lines (b) and (c). But, on Line (d), Chukuemeka wrote the sides the other way around. So, the mistake is on Line (d).

NOTE: Chukuemeka substituted in the values for

\frac{L J}{L K}

$\dfrac{LJ}{LK}$ , which is why he got the correct answer for the proportionality constant. But, this substitution did not match the labels he had written.

STEP: Correct the proof
[−2 points ⇒ 0 / 3 points left]

Chukuemeka should have divided the sides of $\Delta LJM$ by the sides of $\Delta LKJ$ every time.

The correct proof is:

Line	Proof
(a)	In $\Delta LJM$ and $\Delta LKJ$ :
(b)	1. $\dfrac{MJ}{JK} = \dfrac{\text{20}}{\text{12}} = \dfrac{\text{5}}{\text{3}}$
(c)	2. $\dfrac{ML}{JL} = \dfrac{\text{25}}{\text{15}} = \dfrac{\text{5}}{\text{3}}$
(d)	3. $\color{ForestGreen}{\dfrac{LJ}{LK} = \dfrac{\text{15}}{\text{9}} = \dfrac{\text{5}}{\text{3}}}$
(e)	$\therefore \Delta LJM \;\|\|\|\; \Delta LKJ$ (sides of $\Delta$ in prop)

Submit your answer as: and

3.

\frac{L J}{L K} = \frac{15}{9} = \frac{5}{3}

$\dfrac{LJ}{LK} = \dfrac{\text{15}}{\text{9}} = \dfrac{\text{5}}{\text{3}}$

Prove similarity in adjacent triangles

Consider the following diagram:

Prove that the triangles are similar.

Khayone has already answered the question, and his proof is written below. But, Khayone has made a mistake. Look carefully at his proof and identify where he has made his mistake.

Line	Proof
(a)	In $\Delta FEG$ and $\Delta EDG$ :
(b)	1. $\dfrac{EG}{FG} = \dfrac{\text{25}}{\text{15}} = \dfrac{\text{5}}{\text{3}}$
(c)	2. $\dfrac{FE}{ED} = \dfrac{\text{20}}{\text{12}} = \dfrac{\text{5}}{\text{3}}$
(d)	3. $\dfrac{GE}{GD} = \dfrac{\text{15}}{\text{9}} = \dfrac{\text{5}}{\text{3}}$
(e)	$\therefore \Delta FEG \;\|\|\|\; \Delta EDG$ (sides of $\Delta$ in prop)

Answer:

The mistake is on Line .

Replace this line with

$\therefore \Delta FEG \;|||\; \Delta DEG$ (sides of $\Delta$ in prop)
1. $EG$ is common
$\therefore \Delta FEG \equiv \Delta EDG$ (sides of $\Delta$ in prop)
1. $\dfrac{FG}{EG} = \dfrac{\text{25}}{\text{15}} = \dfrac{\text{5}}{\text{3}}$

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Check each line of the proof carefully. Are the sides of the triangles being divided in a consistent order? Is the similarity stated correctly, with the correct reason?

STEP: Identify the error in the proof
[−1 point ⇒ 2 / 3 points left]

When we use sides of $\Delta$ in prop to prove that two triangles are similar, we must divide the sides in the same way each time.

Here, Khayone divided the sides in $\Delta FEG$ by the sides in $\Delta EDG$ on Lines (c) and (d). But, on Line (b), Khayone wrote the sides the other way around. So, the mistake is on Line (b).

NOTE: Khayone substituted in the values for

\frac{F G}{E G}

$\dfrac{FG}{EG}$ , which is why he got the correct answer for the proportionality constant. But, this substitution did not match the labels he had written.

STEP: Correct the proof
[−2 points ⇒ 0 / 3 points left]

Khayone should have divided the sides of $\Delta FEG$ by the sides of $\Delta EDG$ every time.

The correct proof is:

Line	Proof
(a)	In $\Delta FEG$ and $\Delta EDG$ :
(b)	1. $\color{ForestGreen}{\dfrac{FG}{EG} = \dfrac{\text{25}}{\text{15}} = \dfrac{\text{5}}{\text{3}}}$
(c)	2. $\dfrac{FE}{ED} = \dfrac{\text{20}}{\text{12}} = \dfrac{\text{5}}{\text{3}}$
(d)	3. $\dfrac{GE}{GD} = \dfrac{\text{15}}{\text{9}} = \dfrac{\text{5}}{\text{3}}$
(e)	$\therefore \Delta FEG \;\|\|\|\; \Delta EDG$ (sides of $\Delta$ in prop)

Submit your answer as: and

1.

\frac{F G}{E G} = \frac{25}{15} = \frac{5}{3}

$\dfrac{FG}{EG} = \dfrac{\text{25}}{\text{15}} = \dfrac{\text{5}}{\text{3}}$

Prove similarity in adjacent triangles

Consider the following diagram:

Prove that the triangles are similar.

Khadija has already answered the question, and her proof is written below. But, Khadija has made a mistake. Look carefully at her proof and identify where she has made her mistake.

Line	Proof
(a)	In $\Delta LMJ$ and $\Delta JMK$ :
(b)	1. $\dfrac{MJ}{MK} = \dfrac{\text{12}}{\text{9}} = \dfrac{\text{4}}{\text{3}}$
(c)	2. $\dfrac{LM}{JM} = \dfrac{\text{16}}{\text{12}} = \dfrac{\text{4}}{\text{3}}$
(d)	3. $\dfrac{LJ}{JK} = \dfrac{\text{20}}{\text{15}} = \dfrac{\text{4}}{\text{3}}$
(e)	$\therefore \Delta LMJ \;\|\|\|\; \Delta JMK$ (similar $\Delta$ s)

Answer:

The mistake is on Line .

Replace this line with

$\therefore \Delta LMJ \;|||\; \Delta JMK$ (sides of $\Delta$ in prop)
$JM$ is common
2. $\dfrac{JM}{LM} = \dfrac{\text{16}}{\text{12}} = \dfrac{\text{4}}{\text{3}}$
$\therefore \Delta LMJ \equiv \Delta JMK$ (sides of $\Delta$ in prop)

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Check each line of the proof carefully. Are the sides of the triangles being divided in a consistent order? Is the similarity stated correctly, with the correct reason?

STEP: Identify the error in the proof
[−1 point ⇒ 2 / 3 points left]

Khadija has proved that the sides of these two triangles are in proportion. But, on Line (e), Khadija has said that the two triangles were similar because of similar $\Delta$ s. This is not the geometry reason Khadija used to prove similarity. So, the mistake is on Line (e).

TIP: In geometry, we have to use exact reasons.

STEP: Correct the proof
[−2 points ⇒ 0 / 3 points left]

Khadija should have used the correct reason: sides of $\Delta$ in prop.

The correct proof is:

Line	Proof
(a)	In $\Delta LMJ$ and $\Delta JMK$ :
(b)	1. $\dfrac{MJ}{MK} = \dfrac{\text{12}}{\text{9}} = \dfrac{\text{4}}{\text{3}}$
(c)	2. $\dfrac{LM}{JM} = \dfrac{\text{16}}{\text{12}} = \dfrac{\text{4}}{\text{3}}$
(d)	3. $\dfrac{LJ}{JK} = \dfrac{\text{20}}{\text{15}} = \dfrac{\text{4}}{\text{3}}$
(e)	$\color{ForestGreen}{\therefore \Delta LMJ \;\|\|\|\; \Delta JMK}$ (sides of $\color{ForestGreen}{\Delta}$ in prop)

Submit your answer as: and

∴ Δ L M J | | | Δ J M K

$\therefore \Delta LMJ \;|||\; \Delta JMK$ (sides of

Δ

$\Delta$ in prop)

Consequences of similarity

In the diagram below, $\Delta DEF \;|||\; \Delta JLK$ .

Determine the size of $\hat{L}$ .

Answer:

\hat{L} =

$\hat{L} =$

°

$°$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two triangles are similar, then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

STEP: Match up the sides and angles
[−2 points ⇒ 0 / 2 points left]

If two triangles are similar then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

We know that the triangles are similar, so they must be equiangular. We use this fact to determine the size of $\hat{L}$ .

The angles in the second triangle must match the order of the angles in the first triangle, according to the similarity statement $\Delta DEF \;|||\; \Delta JLK$ . ${\color{red}{D}}$ and ${\color{red}{J}}$ are the first letters, so they are a matching pair. ${\color{blue}{E}}$ and ${\color{blue}{L}}$ are the second letters, so they are a matching pair. ${\color{ForestGreen}{F}}$ and ${\color{ForestGreen}{K}}$ are the last letters, so they are a matching pair.

Since $\hat{E}$ and $\hat{L}$ match up, they must be equal. This is only true because we were told that $\Delta DEF \;|||\; \Delta JLK$ . So, we use this as our reason.

\begin{aligned} ∴ \hat{L} & = 48 ° (Δ D E F | | | Δ J L K) \end{aligned}

$\begin{aligned} \therefore \hat{L} &= 48° \;\; (\Delta DEF \;|||\; \Delta JLK) \end{aligned}$

Submit your answer as: and

Consequences of similarity

In the diagram below, $\Delta QRS \;|||\; \Delta ZYX$ .

Determine the length of $ZX$ .

Answer:

Z X =

$ZX =$

units

$\text{units}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two triangles are similar, then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

STEP: Match up the sides and angles
[−2 points ⇒ 0 / 2 points left]

If two triangles are similar then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

We know that the triangles are similar, so their sides will be in proportion. We use this fact to determine the length of $ZX$ .

First we match up the sides. In the statement $\Delta QRS \;|||\; \Delta ZYX$ , ${\color{red}{QR}}$ and ${\color{red}{ZY}}$ are the first two letters, so their sides must match up. Similarly, ${\color{blue}{RS}}$ and ${\color{blue}{YX}}$ match up because their letters came last. Finally, ${\color{ForestGreen}{QS}}$ and ${\color{ForestGreen}{ZX}}$ must match up, because they are the only sides left.

Since $QR$ and $ZY$ match up, we can work out the proportionality constant:

\begin{aligned} k & = \frac{Z Y}{Q R} \\ ∴ k & = \frac{54}{24} \\ ∴ k & = 2,25 \end{aligned}

$\begin{aligned} k & = \dfrac{ZY}{QR} \\ \therefore k & = \dfrac{54}{24} \\ \therefore k & = \text{2,25} \end{aligned}$

This is the number by which all sides in $\Delta QRS$ should by multiplied in order to get the matching sides in $\Delta ZYX$ , because $24 \times \text{2,25} = 54$ .

This is only true because we were told that $\Delta QRS \;|||\; \Delta ZYX$ . So, we use this as our reason.

So,

\begin{aligned} Z X & = 16 \times 2,25 \\ ∴ Z X & = 36 units (Δ Q R S | | | Δ Z Y X) \end{aligned}

$\begin{aligned} ZX & = 16 \times \text{2,25} \\ \therefore ZX & = \text{36}\ \text{units} \;\; (\Delta QRS \;|||\; \Delta ZYX) \end{aligned}$

Submit your answer as: and

Consequences of similarity

In the diagram below, $\Delta MNP \;|||\; \Delta ZXY$ .

Determine the length of $ZY$ .

Answer:

Z Y =

$ZY =$

units

$\text{units}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two triangles are similar, then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

STEP: Match up the sides and angles
[−2 points ⇒ 0 / 2 points left]

If two triangles are similar then

they are equiangular (their matching angles are equal), and
their sides are in proportion (you always have to multiply by the same number to get the matching side in the other triangle).

We know that the triangles are similar, so their sides will be in proportion. We use this fact to determine the length of $ZY$ .

First we match up the sides. In the statement $\Delta MNP \;|||\; \Delta ZXY$ , ${\color{red}{MN}}$ and ${\color{red}{ZX}}$ are the first two letters, so their sides must match up. Similarly, ${\color{blue}{NP}}$ and ${\color{blue}{XY}}$ match up because their letters came last. Finally, ${\color{ForestGreen}{MP}}$ and ${\color{ForestGreen}{ZY}}$ must match up, because they are the only sides left.

Since $MN$ and $ZX$ match up, we can work out the proportionality constant:

\begin{aligned} k & = \frac{Z X}{M N} \\ ∴ k & = \frac{70}{40} \\ ∴ k & = 1,75 \end{aligned}

$\begin{aligned} k & = \dfrac{ZX}{MN} \\ \therefore k & = \dfrac{70}{40} \\ \therefore k & = \text{1,75} \end{aligned}$

This is the number by which all sides in $\Delta MNP$ should by multiplied in order to get the matching sides in $\Delta ZXY$ , because $40 \times \text{1,75} = 70$ .

This is only true because we were told that $\Delta MNP \;|||\; \Delta ZXY$ . So, we use this as our reason.

So,

\begin{aligned} Z Y & = 36 \times 1,75 \\ ∴ Z Y & = 63 units (Δ M N P | | | Δ Z X Y) \end{aligned}

$\begin{aligned} ZY & = 36 \times \text{1,75} \\ \therefore ZY & = \text{63}\ \text{units} \;\; (\Delta MNP \;|||\; \Delta ZXY) \end{aligned}$

Submit your answer as: and

Identify similar triangles

In the diagram below, $TS \bot SP$ and $TP \bot SR$ . Also, $\hat{P} = x$ , $SP = 8$ , and $RP = 5$ .

Jabulile needs to prove that $\Delta TSP \;|||\; \Delta SRP$ . She has already started, and her incomplete proof is written below:

In $\Delta TSP$ and $\Delta SRP$ :
1. $\hat{P}$ is common
2. $T\hat{S}P = S\hat{R}P = 90°$ (given)
  ...
How should Jabulile complete the proof? Choose the best option.
Answer:
3. $\dfrac{SP}{RP} = \dfrac{8}{5}$

$\therefore \Delta TSP \;|||\; \Delta SRP$ (sides of $\Delta$ in prop)

3. $\hat{T} = 180° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $P\hat{S}R = 180° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = P\hat{S}R$ (both equal to $180° - x$ )

$\therefore \Delta TSP \;|||\; \Delta SRP$ (equiangular $\Delta$ s)

3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $P\hat{S}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = P\hat{S}R$ (both equal to $90° - x$ )

$\therefore \Delta TSP \;|||\; \Delta SRP$ (equiangular $\Delta$ s)

3. $P\hat{S}R = 45°$ (given)
$\therefore \hat{T} = 45°$ ( $\Delta TSP \;|||\; \Delta SRP$ )
$\therefore \hat{T} = T\hat{S}R$ (both equal to $45°$ )

$\therefore \Delta TSP \;|||\; \Delta SRP$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Which can you use here?

STEP: Choose the correct proof
[−3 points ⇒ 0 / 3 points left]
NOTE:
There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Jabulile has already proved two pairs of angles equal for us, and we need to prove the third pair. To do so, it is helpful to look at each triangle separately.

Looking at $\Delta TSP$ , we can write $\hat{T}$ in terms of $x$ :

$\begin{aligned} \hat{T} + x + 90 ° & = 180 ° (sum of ∠ s in Δ) \\ ∴ \hat{T} + x + 90 ° - 90 ° & = 180 ° - 90 ° \\ ∴ \hat{T} + x & = 90 ° \\ ∴ \hat{T} + x - x & = 90 ° - x \\ ∴ \hat{T} & = 90 ° - x \end{aligned}$ $\begin{aligned} {\color{ForestGreen}{\hat{T}}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore \hat{T} + x + 90° - 90° & = 18 0° - 90° \\ \therefore \hat{T} + x & = 90° \\ \therefore \hat{T} + x - x & = 90° - x \\ \therefore \hat{T} & = 90° - x \end{aligned}$

In the same way in $\Delta SRP$ , we can write $P\hat{S}R$ in terms of $x$ .

$\begin{aligned} P \hat{S} R + x + 90 ° & = 180 ° (sum of ∠ s in Δ) \\ ∴ P \hat{S} R & = 90 ° - x \end{aligned}$ $\begin{aligned} {\color{ForestGreen}{P\hat{S}R}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore P\hat{S}R & = 90° - x \end{aligned}$

NOTE: We cannot assume that $RS$ cuts $T\hat{S}P$ in half. $P\hat{S}R$ might be equal to $45°$ , or it might be something else. We must only use the information that we have been given.

The full proof is shown below:

In $\color{gray}{\Delta TSP}$ and $\color{gray}{\Delta SRP}$ :
1. $\color{gray}{\hat{P}}$ is common
2. $\color{gray}{T\hat{S}P = S\hat{R}P = 90°}$ (given)
3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  Also, $P\hat{S}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{T} = P\hat{S}R$ (both equal to $90° - x$ )
$\therefore \Delta TSP \;|||\; \Delta SRP$ (equiangular $\Delta$ s)

Submit your answer as:

3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $P\hat{S}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = P\hat{S}R$ (both equal to $90° - x$ )

$\therefore \Delta TSP \;|||\; \Delta SRP$ (equiangular $\Delta$ s)
Hence, determine the length of $TR$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer: $TR =$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $TR$ to the sides that you already know. Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

We are trying to find $TR$ . Even though $TR$ is not a side in the similar triangles, it is part of side $TP$ :

$\begin{matrix} T P = T R + R P \end{matrix}$ $\begin{array}{c} TP = TR + RP \end{array}$

To make the algebra easier, let $TR = y$ .

Now we have

$\begin{matrix} T P = y + 5 \end{matrix}$ $\begin{array}{c} TP = y + 5 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{TP}$ matches with $\color{blue}{SP}$ : $\Delta {\color{blue}{T}}S{\color{blue}{P}} \;|||\; \Delta {\color{blue}{S}}R{\color{blue}{P}}$ .

We know the lengths of ${\color{ForestGreen}{SP}}$ and ${\color{ForestGreen}{RP}}$ . They are a matching pair of sides: ( $\Delta T{\color{ForestGreen}{S}}{\color{ForestGreen}{P}} \;|||\; \Delta S{\color{ForestGreen}{R}}{\color{ForestGreen}{P}}$ ).

So, dividing the sides of the big triangle by the sides of the small triangle, we get:

$\begin{aligned} \frac{T P}{S P} & = \frac{S P}{R P} (Δ T S P | | | Δ S R P) \end{aligned}$ $\begin{aligned} \dfrac{TP}{SP} & = \dfrac{SP}{RP} \;\; (\Delta TSP \;|||\; \Delta SRP) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

$\begin{aligned} \frac{y + 5}{8} & = \frac{8}{5} \\ \frac{y + 5}{8} \times 8 & = \frac{8}{5} \times 8 \\ y + 5 & = \frac{64}{5} \\ y + 5 - 5 & = \frac{64}{5} - 5 \\ ∴ y & = \frac{39}{5} \end{aligned}$ $\begin{aligned} {\color{blue}{\dfrac{y + 5}{8}}} & = {\color{ForestGreen}{\dfrac{8}{5}}} \\ \dfrac{y + 5}{8} \times 8 & = \dfrac{8}{5} \times 8 \\ y + 5 & = \dfrac{64}{5} \\ y + 5 - 5 & = \dfrac{64}{5} - 5 \\ \therefore y & = \frac{39}{5} \end{aligned}$

Submit your answer as: and

Identify similar triangles

In the diagram below, $CE \bot EA$ and $CA \bot ED$ . Also, $\hat{A} = x$ , $EA = 6$ , and $DA = 5$ .

Khayalethu needs to prove that $\Delta CEA \;|||\; \Delta EDA$ . He has already started, and his incomplete proof is written below:

In $\Delta CEA$ and $\Delta EDA$ :
1. $\hat{A}$ is common
2. $C\hat{E}A = E\hat{D}A = 90°$ (given)
  ...
How should Khayalethu complete the proof? Choose the best option.
Answer:
3. $\hat{C} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $A\hat{E}D = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = A\hat{E}D$ (both equal to $90° - x$ )

$\therefore \Delta CEA \;|||\; \Delta EDA$ (equiangular $\Delta$ s)

3. $\dfrac{EA}{DA} = \dfrac{6}{5}$

$\therefore \Delta CEA \;|||\; \Delta EDA$ (sides of $\Delta$ in prop)

3. $\hat{C} = 180° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $A\hat{E}D = 180° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = A\hat{E}D$ (both equal to $180° - x$ )

$\therefore \Delta CEA \;|||\; \Delta EDA$ (equiangular $\Delta$ s)

3. $A\hat{E}D = 45°$ (given)
$\therefore \hat{C} = 45°$ ( $\Delta CEA \;|||\; \Delta EDA$ )
$\therefore \hat{C} = C\hat{E}D$ (both equal to $45°$ )

$\therefore \Delta CEA \;|||\; \Delta EDA$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Which can you use here?

STEP: Choose the correct proof
[−3 points ⇒ 0 / 3 points left]
NOTE:
There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Khayalethu has already proved two pairs of angles equal for us, and we need to prove the third pair. To do so, it is helpful to look at each triangle separately.

Looking at $\Delta CEA$ , we can write $\hat{C}$ in terms of $x$ :

$\begin{aligned} \hat{C} + x + 90 ° & = 180 ° (sum of ∠ s in Δ) \\ ∴ \hat{C} + x + 90 ° - 90 ° & = 180 ° - 90 ° \\ ∴ \hat{C} + x & = 90 ° \\ ∴ \hat{C} + x - x & = 90 ° - x \\ ∴ \hat{C} & = 90 ° - x \end{aligned}$ $\begin{aligned} {\color{ForestGreen}{\hat{C}}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore \hat{C} + x + 90° - 90° & = 18 0° - 90° \\ \therefore \hat{C} + x & = 90° \\ \therefore \hat{C} + x - x & = 90° - x \\ \therefore \hat{C} & = 90° - x \end{aligned}$

In the same way in $\Delta EDA$ , we can write $A\hat{E}D$ in terms of $x$ .

$\begin{aligned} A \hat{E} D + x + 90 ° & = 180 ° (sum of ∠ s in Δ) \\ ∴ A \hat{E} D & = 90 ° - x \end{aligned}$ $\begin{aligned} {\color{ForestGreen}{A\hat{E}D}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore A\hat{E}D & = 90° - x \end{aligned}$

NOTE: We cannot assume that $DE$ cuts $C\hat{E}A$ in half. $A\hat{E}D$ might be equal to $45°$ , or it might be something else. We must only use the information that we have been given.

The full proof is shown below:

In $\color{gray}{\Delta CEA}$ and $\color{gray}{\Delta EDA}$ :
1. $\color{gray}{\hat{A}}$ is common
2. $\color{gray}{C\hat{E}A = E\hat{D}A = 90°}$ (given)
3. $\hat{C} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  Also, $A\hat{E}D = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{C} = A\hat{E}D$ (both equal to $90° - x$ )
$\therefore \Delta CEA \;|||\; \Delta EDA$ (equiangular $\Delta$ s)

Submit your answer as:

3. $\hat{C} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $A\hat{E}D = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{C} = A\hat{E}D$ (both equal to $90° - x$ )

$\therefore \Delta CEA \;|||\; \Delta EDA$ (equiangular $\Delta$ s)
Hence, determine the length of $CD$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer: $CD =$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $CD$ to the sides that you already know. Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

We are trying to find $CD$ . Even though $CD$ is not a side in the similar triangles, it is part of side $CA$ :

$\begin{matrix} C A = C D + D A \end{matrix}$ $\begin{array}{c} CA = CD + DA \end{array}$

To make the algebra easier, let $CD = y$ .

Now we have

$\begin{matrix} C A = y + 5 \end{matrix}$ $\begin{array}{c} CA = y + 5 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{CA}$ matches with $\color{blue}{EA}$ : $\Delta {\color{blue}{C}}E{\color{blue}{A}} \;|||\; \Delta {\color{blue}{E}}D{\color{blue}{A}}$ .

We know the lengths of ${\color{ForestGreen}{EA}}$ and ${\color{ForestGreen}{DA}}$ . They are a matching pair of sides: ( $\Delta C{\color{ForestGreen}{E}}{\color{ForestGreen}{A}} \;|||\; \Delta E{\color{ForestGreen}{D}}{\color{ForestGreen}{A}}$ ).

So, dividing the sides of the big triangle by the sides of the small triangle, we get:

$\begin{aligned} \frac{C A}{E A} & = \frac{E A}{D A} (Δ C E A | | | Δ E D A) \end{aligned}$ $\begin{aligned} \dfrac{CA}{EA} & = \dfrac{EA}{DA} \;\; (\Delta CEA \;|||\; \Delta EDA) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

$\begin{aligned} \frac{y + 5}{6} & = \frac{6}{5} \\ \frac{y + 5}{6} \times 6 & = \frac{6}{5} \times 6 \\ y + 5 & = \frac{36}{5} \\ y + 5 - 5 & = \frac{36}{5} - 5 \\ ∴ y & = \frac{11}{5} \end{aligned}$ $\begin{aligned} {\color{blue}{\dfrac{y + 5}{6}}} & = {\color{ForestGreen}{\dfrac{6}{5}}} \\ \dfrac{y + 5}{6} \times 6 & = \dfrac{6}{5} \times 6 \\ y + 5 & = \dfrac{36}{5} \\ y + 5 - 5 & = \dfrac{36}{5} - 5 \\ \therefore y & = \frac{11}{5} \end{aligned}$

Submit your answer as: and

Identify similar triangles

In the diagram below, $TQ \bot QP$ and $TP \bot QR$ . Also, $\hat{P} = x$ , $QP = 8$ , and $RP = 5$ .

Daniel needs to prove that $\Delta TQP \;|||\; \Delta QRP$ . He has already started, and his incomplete proof is written below:

In $\Delta TQP$ and $\Delta QRP$ :
1. $\hat{P}$ is common
2. $T\hat{Q}P = Q\hat{R}P = 90°$ (given)
  ...
How should Daniel complete the proof? Choose the best option.
Answer:
3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $P\hat{Q}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = P\hat{Q}R$ (both equal to $90° - x$ )

$\therefore \Delta TQP \;|||\; \Delta QRP$ (equiangular $\Delta$ s)

3. $\dfrac{QP}{RP} = \dfrac{8}{5}$

$\therefore \Delta TQP \;|||\; \Delta QRP$ (sides of $\Delta$ in prop)

3. $\hat{T} = 180° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $P\hat{Q}R = 180° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = P\hat{Q}R$ (both equal to $180° - x$ )

$\therefore \Delta TQP \;|||\; \Delta QRP$ (equiangular $\Delta$ s)

3. $P\hat{Q}R = 45°$ (given)
$\therefore \hat{T} = 45°$ ( $\Delta TQP \;|||\; \Delta QRP$ )
$\therefore \hat{T} = T\hat{Q}R$ (both equal to $45°$ )

$\therefore \Delta TQP \;|||\; \Delta QRP$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Which can you use here?

STEP: Choose the correct proof
[−3 points ⇒ 0 / 3 points left]
NOTE:
There are two ways of proving similarity in triangles:
- Prove that the two triangles are equiangular.
- Prove that the two triangles have matching sides in proportion.
Daniel has already proved two pairs of angles equal for us, and we need to prove the third pair. To do so, it is helpful to look at each triangle separately.

Looking at $\Delta TQP$ , we can write $\hat{T}$ in terms of $x$ :

$\begin{aligned} \hat{T} + x + 90 ° & = 180 ° (sum of ∠ s in Δ) \\ ∴ \hat{T} + x + 90 ° - 90 ° & = 180 ° - 90 ° \\ ∴ \hat{T} + x & = 90 ° \\ ∴ \hat{T} + x - x & = 90 ° - x \\ ∴ \hat{T} & = 90 ° - x \end{aligned}$ $\begin{aligned} {\color{ForestGreen}{\hat{T}}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore \hat{T} + x + 90° - 90° & = 18 0° - 90° \\ \therefore \hat{T} + x & = 90° \\ \therefore \hat{T} + x - x & = 90° - x \\ \therefore \hat{T} & = 90° - x \end{aligned}$

In the same way in $\Delta QRP$ , we can write $P\hat{Q}R$ in terms of $x$ .

$\begin{aligned} P \hat{Q} R + x + 90 ° & = 180 ° (sum of ∠ s in Δ) \\ ∴ P \hat{Q} R & = 90 ° - x \end{aligned}$ $\begin{aligned} {\color{ForestGreen}{P\hat{Q}R}} + {\color{blue}{x}} + {\color{red}{90°}} & = 180° \;\; (\text{sum of } \angle \text{s in } \Delta) \\ \therefore P\hat{Q}R & = 90° - x \end{aligned}$

NOTE: We cannot assume that $RQ$ cuts $T\hat{Q}P$ in half. $P\hat{Q}R$ might be equal to $45°$ , or it might be something else. We must only use the information that we have been given.

The full proof is shown below:

In $\color{gray}{\Delta TQP}$ and $\color{gray}{\Delta QRP}$ :
1. $\color{gray}{\hat{P}}$ is common
2. $\color{gray}{T\hat{Q}P = Q\hat{R}P = 90°}$ (given)
3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  Also, $P\hat{Q}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{T} = P\hat{Q}R$ (both equal to $90° - x$ )
$\therefore \Delta TQP \;|||\; \Delta QRP$ (equiangular $\Delta$ s)

Submit your answer as:

3. $\hat{T} = 90° - x$ (sum of $\angle$ s in $\Delta$ )
Also, $P\hat{Q}R = 90° - x$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = P\hat{Q}R$ (both equal to $90° - x$ )

$\therefore \Delta TQP \;|||\; \Delta QRP$ (equiangular $\Delta$ s)
Hence, determine the length of $TR$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer: $TR =$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $TR$ to the sides that you already know. Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

We are trying to find $TR$ . Even though $TR$ is not a side in the similar triangles, it is part of side $TP$ :

$\begin{matrix} T P = T R + R P \end{matrix}$ $\begin{array}{c} TP = TR + RP \end{array}$

To make the algebra easier, let $TR = y$ .

Now we have

$\begin{matrix} T P = y + 5 \end{matrix}$ $\begin{array}{c} TP = y + 5 \end{array}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{TP}$ matches with $\color{blue}{QP}$ : $\Delta {\color{blue}{T}}Q{\color{blue}{P}} \;|||\; \Delta {\color{blue}{Q}}R{\color{blue}{P}}$ .

We know the lengths of ${\color{ForestGreen}{QP}}$ and ${\color{ForestGreen}{RP}}$ . They are a matching pair of sides: ( $\Delta T{\color{ForestGreen}{Q}}{\color{ForestGreen}{P}} \;|||\; \Delta Q{\color{ForestGreen}{R}}{\color{ForestGreen}{P}}$ ).

So, dividing the sides of the big triangle by the sides of the small triangle, we get:

$\begin{aligned} \frac{T P}{Q P} & = \frac{Q P}{R P} (Δ T Q P | | | Δ Q R P) \end{aligned}$ $\begin{aligned} \dfrac{TP}{QP} & = \dfrac{QP}{RP} \;\; (\Delta TQP \;|||\; \Delta QRP) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

$\begin{aligned} \frac{y + 5}{8} & = \frac{8}{5} \\ \frac{y + 5}{8} \times 8 & = \frac{8}{5} \times 8 \\ y + 5 & = \frac{64}{5} \\ y + 5 - 5 & = \frac{64}{5} - 5 \\ ∴ y & = \frac{39}{5} \end{aligned}$ $\begin{aligned} {\color{blue}{\dfrac{y + 5}{8}}} & = {\color{ForestGreen}{\dfrac{8}{5}}} \\ \dfrac{y + 5}{8} \times 8 & = \dfrac{8}{5} \times 8 \\ y + 5 & = \dfrac{64}{5} \\ y + 5 - 5 & = \dfrac{64}{5} - 5 \\ \therefore y & = \frac{39}{5} \end{aligned}$

Submit your answer as: and

Definition: Similarity

Two triangles are similar if:

Answer:

some of their sides are the same length.
they are exactly the same shape, but they may be different sizes.
they are approximately the same shape.
some of their angles are equal.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

What do we mean in maths when we say that two triangles are similar?

STEP: Identify the definition of similarity
[−2 points ⇒ 0 / 2 points left]

We say that two triangles are similar if they are exactly the same shape, but one might be bigger or smaller than the other one. Another way of saying this is that one is an enlargement of the other one.

For example, $\Delta DEF$ and $\Delta JKL$ below are the same shape. But, they are different sizes.

The sides of similar triangles are in proportion. That means that all of the sides in one triangle have been multiplied by the same number to get the sides in the second triangle. All of the sides in $\Delta DEF$ have been multiplied by $\text{0,5}$ to get $\Delta JKL$ .

Similar triangles are the same shape, so they always have the same angles. We can see the matching angles in $\Delta DEF$ and $\Delta JKL$ .

NOTE: In normal English, we say two things are similar if they are almost the same. But, in Maths, it is not precise enough to say that similar shapes are approximately the same. In Maths, similarity has a very specific meaning.

Two triangles are similar if either

their matching sides are in proportion, or
their matching angles are equal.

NOTE: In many polygons, we have to prove that the matching sides are in proportion and the matching angles are equal, if we want to be sure that the shapes are similar. But, in triangles, we only need to prove that one of them is true. This is a special property of triangles, which you will learn more about in grade

12

$\text{12}$ .

Submit your answer as:

they are exactly the same shape, but they may be different sizes.

Definition: Similarity

Two triangles are similar if:

Answer:

they are exactly the same shape, but they may be different sizes.
their angles are in proportion.
some of their angles are equal.
they are approximately the same shape.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

What do we mean in maths when we say that two triangles are similar?

STEP: Identify the definition of similarity
[−2 points ⇒ 0 / 2 points left]

We say that two triangles are similar if they are exactly the same shape, but one might be bigger or smaller than the other one. Another way of saying this is that one is an enlargement of the other one.

For example, $\Delta ABC$ and $\Delta DEF$ below are the same shape. But, they are different sizes.

The sides of similar triangles are in proportion. That means that all of the sides in one triangle have been multiplied by the same number to get the sides in the second triangle. All of the sides in $\Delta ABC$ have been multiplied by $\text{0,5}$ to get $\Delta DEF$ .

Similar triangles are the same shape, so they always have the same angles. We can see the matching angles in $\Delta ABC$ and $\Delta DEF$ .

NOTE: In normal English, we say two things are similar if they are almost the same. But, in Maths, it is not precise enough to say that similar shapes are approximately the same. In Maths, similarity has a very specific meaning.

Two triangles are similar if either

their matching sides are in proportion, or
their matching angles are equal.

NOTE: In many polygons, we have to prove that the matching sides are in proportion and the matching angles are equal, if we want to be sure that the shapes are similar. But, in triangles, we only need to prove that one of them is true. This is a special property of triangles, which you will learn more about in grade

12

$\text{12}$ .

Submit your answer as:

they are exactly the same shape, but they may be different sizes.

Definition: Similarity

Two triangles are similar if:

Answer:

some of their angles are equal.
they are approximately the same shape.
their matching sides are in proportion.
some of their sides are the same length.

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

What do we mean in maths when we say that two triangles are similar?

STEP: Identify the definition of similarity
[−2 points ⇒ 0 / 2 points left]

We say that two triangles are similar if they are exactly the same shape, but one might be bigger or smaller than the other one. Another way of saying this is that one is an enlargement of the other one.

For example, $\Delta PQR$ and $\Delta XYZ$ below are the same shape. But, they are different sizes.

The sides of similar triangles are in proportion. That means that all of the sides in one triangle have been multiplied by the same number to get the sides in the second triangle. All of the sides in $\Delta PQR$ have been multiplied by $\text{0,5}$ to get $\Delta XYZ$ .

Similar triangles are the same shape, so they always have the same angles. We can see the matching angles in $\Delta PQR$ and $\Delta XYZ$ .

NOTE: In normal English, we say two things are similar if they are almost the same. But, in Maths, it is not precise enough to say that similar shapes are approximately the same. In Maths, similarity has a very specific meaning.

Two triangles are similar if either

their matching sides are in proportion, or
their matching angles are equal.

NOTE: In many polygons, we have to prove that the matching sides are in proportion and the matching angles are equal, if we want to be sure that the shapes are similar. But, in triangles, we only need to prove that one of them is true. This is a special property of triangles, which you will learn more about in grade

12

$\text{12}$ .

Submit your answer as:

their matching sides are in proportion.

Using the proportionality of similar polygons

In the diagram below, $RSXY\; ||| \;$ $TWYZ$ . Compute the value of $f$ . The diagram is not necessarily drawn to scale.

Answer:

f =

$f =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Similar shapes have proportional lengths. Write a proportion for the sides of the polygons and then solve the equation for $f$ .

STEP: Identify corresponding sides and write a proportion for these sides
[−1 point ⇒ 2 / 3 points left]

We have a diagram with quadrilaterals, and we need to figure out the value of the variable $f$ . The question tells us that the quadrilaterals are similar, which means that the sides of the shapes are proportional. Therefore, we can write a proportion for the sides of the polygons and solve it for $f$ .

There are a number of different proportions that we can write to summarise the lengths of the sides in the diagram: we can write a proportion for any pairs of corresponding sides. Here are two proportions which are accurate (but there are others):

\begin{matrix} \frac{R S}{T Z} = \frac{X Y}{W Y} \frac{R S}{X Y} = \frac{T Z}{W Y} \end{matrix}

$\begin{array}{c}\frac{RS}{TZ} = \frac{XY}{WY} \qquad \frac{RS}{XY} = \frac{TZ}{WY}\end{array}$

STEP: Solve the equation for

f

$f$
[−2 points ⇒ 0 / 3 points left]

We will use the first proportion from above to find the value of $f$ . If you use a different correct proportion and solve it without any mistakes, you will get the correct answer for $f$ (and full marks).

Remember:

\begin{aligned} W Y & = X Y + W X \\ = - 2 f + 3 + 4 \\ = - 2 f + 7 \end{aligned}

$\begin{aligned} WY &= XY + WX \\ &= - 2 f + 3 + 4 \\ &= - 2 f + 7 \end{aligned}$

So now we can calculate $f$ :

\begin{aligned} \frac{R S}{T Z} & = \frac{X Y}{W Y} \\ \frac{2 f}{2 f + 12} & = \frac{- 2 f + 3}{- 2 f + 7} \\ (2 f) (- 2 f + 7) & = (- 2 f + 3) (2 f + 12) \\ - 4 f^{2} + 14 f & = - 4 f^{2} - 18 f + 36 ⟵_{cancel}^{- 4 f^{2} terms} \\ 32 f & = 36 \\ f & = \frac{36}{32} \\ f & = \frac{9}{8} \end{aligned}

$\begin{aligned} \frac{RS}{TZ} & = \frac{XY}{WY} \\ \frac{2 f}{2 f + 12} & = \frac{- 2 f + 3}{- 2 f + 7} \\ \left( 2 f \right) \left( - 2 f + 7 \right) & = \left(- 2 f + 3 \right) \left( 2 f + 12 \right) \\ - 4 f^{2} + 14 f & = - 4 f^{2} - 18 f + 36 \qquad \longleftarrow ^{- 4 f^{2} \text{ terms}}_\text{cancel} \\ 32 f&= 36 \\ f &= \frac{36}{32} \\ f & = \frac{9}{8} \end{aligned}$

Therefore, the value of the variable is $f = \frac{9}{8}$ .

Submit your answer as:

Using the proportionality of similar polygons

Triangles $PQT$ and $PRS$ are similar. Compute the value of $g$ . The diagram is not necessarily drawn to scale.

Answer:

g =

$g =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Similar shapes have proportional lengths. Write a proportion for the sides of the polygons and then solve the equation for $g$ .

STEP: Identify corresponding sides and write a proportion for these sides
[−1 point ⇒ 2 / 3 points left]

We have a diagram with triangles, and we need to figure out the value of the variable $g$ . The question tells us that the triangles are similar, which means that the sides of the shapes are proportional. Therefore, we can write a proportion for the sides of the polygons and solve it for $g$ .

There are a number of different proportions that we can write to summarise the lengths of the sides in the diagram: we can write a proportion for any pairs of corresponding sides. Here are two proportions which are accurate (but there are others):

\begin{matrix} \frac{P Q}{Q R} = \frac{P T}{S T} \frac{P Q}{P R} = \frac{P T}{P S} \end{matrix}

$\begin{array}{c}\frac{PQ}{QR} = \frac{PT}{ST} \qquad \frac{PQ}{PR} = \frac{PT}{PS}\end{array}$

STEP: Solve the equation for

g

$g$
[−2 points ⇒ 0 / 3 points left]

We will use the first proportion from above to find the value of $g$ . If you use a different correct proportion and solve it without any mistakes, you will get the correct answer for $g$ (and full marks).

\begin{aligned} \frac{P Q}{Q R} & = \frac{P T}{S T} \\ \frac{2 g}{15} & = \frac{- 3 g + 4}{5} \\ (2 g) (5) & = (- 3 g + 4) (15) \\ 10 g & = - 45 g + 60 \\ 55 g & = 60 \\ g & = \frac{60}{55} \\ g & = \frac{12}{11} \end{aligned}

$\begin{aligned} \frac{PQ}{QR} & = \frac{PT}{ST} \\ \frac{2 g}{15} & = \frac{- 3 g + 4}{5} \\ \left( 2 g \right) \left( 5 \right) & = \left(- 3 g + 4 \right) \left( 15 \right) \\ 10 g & = - 45 g + 60 \\ 55 g&= 60 \\ g &= \frac{60}{55} \\ g & = \frac{12}{11} \end{aligned}$

Therefore, the value of the variable is $g = \frac{12}{11}$ .

Submit your answer as:

Using the proportionality of similar polygons

Triangles $PQT$ and $PRS$ are similar. What is the value of $f$ ? The diagram is not necessarily drawn to scale.

Answer:

f =

$f =$

numeric

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Similar shapes have proportional lengths. Write a proportion for the sides of the polygons and then solve the equation for $f$ .

STEP: Identify corresponding sides and write a proportion for these sides
[−1 point ⇒ 2 / 3 points left]

We have a diagram with triangles, and we need to figure out the value of the variable $f$ . The question tells us that the triangles are similar, which means that the sides of the shapes are proportional. Therefore, we can write a proportion for the sides of the polygons and solve it for $f$ .

There are a number of different proportions that we can write to summarise the lengths of the sides in the diagram: we can write a proportion for any pairs of corresponding sides. Here are two proportions which are accurate (but there are others):

\begin{matrix} \frac{P Q}{Q R} = \frac{P T}{S T} \frac{P Q}{P R} = \frac{P T}{P S} \end{matrix}

$\begin{array}{c}\frac{PQ}{QR} = \frac{PT}{ST} \qquad \frac{PQ}{PR} = \frac{PT}{PS}\end{array}$

STEP: Solve the equation for

f

$f$
[−2 points ⇒ 0 / 3 points left]

We will use the first proportion from above to find the value of $f$ . If you use a different correct proportion and solve it without any mistakes, you will get the correct answer for $f$ (and full marks).

\begin{aligned} \frac{P Q}{Q R} & = \frac{P T}{S T} \\ \frac{3 f}{3} & = \frac{f + 2}{6} \\ (3 f) (6) & = (f + 2) (3) \\ 18 f & = 3 f + 6 \\ 15 f & = 6 \\ f & = \frac{6}{15} \\ f & = \frac{2}{5} \end{aligned}

$\begin{aligned} \frac{PQ}{QR} & = \frac{PT}{ST} \\ \frac{3 f}{3} & = \frac{f + 2}{6} \\ \left( 3 f \right) \left( 6 \right) & = \left(f + 2 \right) \left( 3 \right) \\ 18 f & = 3 f + 6 \\ 15 f&= 6 \\ f &= \frac{6}{15} \\ f & = \frac{2}{5} \end{aligned}$

Therefore, the value of the variable is $f = \frac{2}{5}$ .

Submit your answer as:

Given similar triangles, calculate a side

In the diagram below, $\Delta BED \;|||\; \Delta CEA$ . Also, $BD = 8$ , $CA = 3$ , $CB = 7$ , and $EC = y$ .

Determine the value of $y$ , giving a reason for your answer.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer:

y =

$y =$

units

$\text{units}$

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $y$ to the sides that you already know. You may have $y$ more than once in your equation!

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

If two triangles are similar, their sides are in proportion. So, when you divide the sides of one triangle with the matching sides in the other triangle, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios by making pairs of matching sides.

We are trying to find $y$ . We can see that $y$ is equal to $EC$ . We can also see that:

\begin{aligned} E B = y + 7 \end{aligned}

$\begin{aligned} EB = y + 7 \end{aligned}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{BE}$ matches with $\color{blue}{CE}$ : $\Delta {\color{blue}{B}}{\color{blue}{E}}D \;|||\; \Delta {\color{blue}{C}}{\color{blue}{E}}A$ .

We know the lengths of $\color{red}{BD}$ and $\color{red}{CA}$ . They are a matching pair of sides: $\Delta {\color{red}{B}}E{\color{red}{D}} \;|||\; \Delta {\color{red}{C}}E{\color{red}{A}}$ .

So, dividing the sides in the big triangle by those in the small triangle, we get

\begin{aligned} \frac{B E}{C E} = \frac{B D}{C A} (Δ B E D | | | Δ C E A) \end{aligned}

$\begin{aligned} \dfrac{BE}{CE} = \dfrac{BD}{CA} \; \left(\Delta BED \;|||\; \Delta CEA\right) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

\begin{aligned} \frac{y + 7}{y} & = \frac{8}{3} \\ ∴ & \frac{y + 7}{y} \times y & = \frac{8}{3} \times y \\ ∴ & y + 7 & = \frac{8 y}{3} \\ ∴ & (y + 7) \times 3 & = \frac{8 y}{3} \times 3 \\ ∴ & 3 y + 21 & = 8 y \\ ∴ & 3 y + 21 - 3 y & = 8 y - 3 y \\ ∴ & 21 & = 5 y \\ ∴ & y & = \frac{21}{5} \end{aligned}

$\begin{alignedat}{3} && \dfrac{y + 7}{y} & = \dfrac{8}{3} \\ \therefore && \dfrac{y + 7}{y} \times y & = \dfrac{8}{3} \times y \\ \therefore && y + 7 &= \dfrac{8y}{3} \\ \therefore && (y + 7) \times 3 & = \dfrac{8y}{3} \times 3 \\ \therefore && 3y + 21 & = 8y \\ \therefore && 3y + 21 - 3y & = 8y - 3y \\ \therefore && 21 & = 5y \\ \therefore && y & = \frac{21}{5} \end{alignedat}$

Submit your answer as: and

Given similar triangles, calculate a side

In the diagram below, $\Delta PTS \;|||\; \Delta PRQ$ . Also, $TS = 8$ , $RQ = 3$ , $RT = 7$ , and $PR = b$ .

Determine the value of $b$ , giving a reason for your answer.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer:

b =

$b =$

units

$\text{units}$

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $b$ to the sides that you already know. You may have $b$ more than once in your equation!

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

If two triangles are similar, their sides are in proportion. So, when you divide the sides of one triangle with the matching sides in the other triangle, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios by making pairs of matching sides.

We are trying to find $b$ . We can see that $b$ is equal to $PR$ . We can also see that:

\begin{aligned} P T = b + 7 \end{aligned}

$\begin{aligned} PT = b + 7 \end{aligned}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{PT}$ matches with $\color{blue}{PR}$ : $\Delta {\color{blue}{P}}{\color{blue}{T}}S \;|||\; \Delta {\color{blue}{P}}{\color{blue}{R}}Q$ .

We know the lengths of $\color{red}{TS}$ and $\color{red}{RQ}$ . They are a matching pair of sides: $\Delta P{\color{red}{T}}{\color{red}{S}} \;|||\; \Delta P{\color{red}{R}}{\color{red}{Q}}$ .

So, dividing the sides in the big triangle by those in the small triangle, we get

\begin{aligned} \frac{P T}{P R} = \frac{T S}{R Q} (Δ P T S | | | Δ P R Q) \end{aligned}

$\begin{aligned} \dfrac{PT}{PR} = \dfrac{TS}{RQ} \; \left(\Delta PTS \;|||\; \Delta PRQ\right) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $b$ -terms, we will be able to solve the equation.

\begin{aligned} \frac{b + 7}{b} & = \frac{8}{3} \\ ∴ & \frac{b + 7}{b} \times b & = \frac{8}{3} \times b \\ ∴ & b + 7 & = \frac{8 b}{3} \\ ∴ & (b + 7) \times 3 & = \frac{8 b}{3} \times 3 \\ ∴ & 3 b + 21 & = 8 b \\ ∴ & 3 b + 21 - 3 b & = 8 b - 3 b \\ ∴ & 21 & = 5 b \\ ∴ & b & = \frac{21}{5} \end{aligned}

$\begin{alignedat}{3} && \dfrac{b + 7}{b} & = \dfrac{8}{3} \\ \therefore && \dfrac{b + 7}{b} \times b & = \dfrac{8}{3} \times b \\ \therefore && b + 7 &= \dfrac{8b}{3} \\ \therefore && (b + 7) \times 3 & = \dfrac{8b}{3} \times 3 \\ \therefore && 3b + 21 & = 8b \\ \therefore && 3b + 21 - 3b & = 8b - 3b \\ \therefore && 21 & = 5b \\ \therefore && b & = \frac{21}{5} \end{alignedat}$

Submit your answer as: and

Given similar triangles, calculate a side

In the diagram below, $\Delta KLJ \;|||\; \Delta MLN$ . Also, $JK = 9$ , $NM = 4$ , $NJ = 7$ , and $LN = y$ .

Determine the value of $y$ , giving a reason for your answer.

INSTRUCTION: Give your answer as a fraction, where necessary.

Answer:

y =

$y =$

units

$\text{units}$

one-of

type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Since the triangles are similar, you know that the sides are in proportion. Use this fact to create an equation which connects $y$ to the sides that you already know. You may have $y$ more than once in your equation!

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 2 / 4 points left]

If two triangles are similar, their sides are in proportion. So, when you divide the sides of one triangle with the matching sides in the other triangle, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios by making pairs of matching sides.

We are trying to find $y$ . We can see that $y$ is equal to $LN$ . We can also see that:

\begin{aligned} L J = y + 7 \end{aligned}

$\begin{aligned} LJ = y + 7 \end{aligned}$

The similarity is easier to see if we separate the triangles.

We can see from the similarity statement that $\color{blue}{LJ}$ matches with $\color{blue}{LN}$ : $\Delta K{\color{blue}{L}}{\color{blue}{J}} \;|||\; \Delta M{\color{blue}{L}}{\color{blue}{N}}$ .

We know the lengths of $\color{red}{KJ}$ and $\color{red}{MN}$ . They are a matching pair of sides: $\Delta {\color{red}{K}}L{\color{red}{J}} \;|||\; \Delta {\color{red}{M}}L{\color{red}{N}}$ .

So, dividing the sides in the big triangle by those in the small triangle, we get

\begin{aligned} \frac{L J}{L N} = \frac{K J}{M N} (Δ K L J | | | Δ M L N) \end{aligned}

$\begin{aligned} \dfrac{LJ}{LN} = \dfrac{KJ}{MN} \; \left(\Delta KLJ \;|||\; \Delta MLN\right) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Solve the equation using inverse operations
[−2 points ⇒ 0 / 4 points left]

First we need to substitute in the values from the diagram. As long as there are only numbers and $y$ -terms, we will be able to solve the equation.

\begin{aligned} \frac{y + 7}{y} & = \frac{9}{4} \\ ∴ & \frac{y + 7}{y} \times y & = \frac{9}{4} \times y \\ ∴ & y + 7 & = \frac{9 y}{4} \\ ∴ & (y + 7) \times 4 & = \frac{9 y}{4} \times 4 \\ ∴ & 4 y + 28 & = 9 y \\ ∴ & 4 y + 28 - 4 y & = 9 y - 4 y \\ ∴ & 28 & = 5 y \\ ∴ & y & = \frac{28}{5} \end{aligned}

$\begin{alignedat}{3} && \dfrac{y + 7}{y} & = \dfrac{9}{4} \\ \therefore && \dfrac{y + 7}{y} \times y & = \dfrac{9}{4} \times y \\ \therefore && y + 7 &= \dfrac{9y}{4} \\ \therefore && (y + 7) \times 4 & = \dfrac{9y}{4} \times 4 \\ \therefore && 4y + 28 & = 9y \\ \therefore && 4y + 28 - 4y & = 9y - 4y \\ \therefore && 28 & = 5y \\ \therefore && y & = \frac{28}{5} \end{alignedat}$

Submit your answer as: and

Given similar triangles, calculate a side

In the diagram below, $\Delta CDA \;|||\; \Delta CBD$ . Also, $BD = 12$ , $CD = 13$ , and $CB = 5$ .

Dumile has been asked to determine the length of $DA$ . He knows that in order to calculate $DA$ , he first needs to match up the sides of the two similar triangles.

Help Dumile to match up the sides in the similar triangles.

Answer:

Side in $\Delta CDA$	Matching side in $\Delta CBD$
$CD$
$CA$
$DA$

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the similarity statement to choose the correct pairs of sides.

STEP: Use the similarity statement to match the sides
[−2 points ⇒ 0 / 2 points left]

Since $\Delta {\color{red}{C}}{\color{red}{D}}A \;|||\; \Delta {\color{red}{C}}{\color{red}{B}}D$ , we can see that ${\color{red}{CD}}$ matches with ${\color{red}{CB}}$ . These are the first two letters in the similarity statement.

Also using $\Delta {\color{blue}{C}}D{\color{blue}{A}} \;|||\; \Delta {\color{blue}{C}}B{\color{blue}{D}}$ , we can see that ${\color{blue}{CA}}$ matches with ${\color{blue}{CD}}$ . These are the first and last letters in the similarity statement.

Finally, we can use $\Delta C{\color{ForestGreen}{D}}{\color{ForestGreen}{A}} \;|||\; \Delta C{\color{ForestGreen}{B}}{\color{ForestGreen}{D}}$ to work out that ${\color{ForestGreen}{DA}}$ matches with ${\color{ForestGreen}{BD}}$ . These are the second and third letters in the similarity statement.

Visually we can see the matching sides of the triangles:

Submit your answer as: andand

Determine the value of $DA$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer: $DA =$ $\text{units}$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similaity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, then their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $DA$ . So, start by writing $DA$ in the numerator:

$\begin{aligned} \frac{D A}{◻} & = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{DA}{\square} & = \dfrac{\square}{\square} \end{aligned}$

Since $\Delta C{\color{ForestGreen}{D}}{\color{ForestGreen}{A}} \;|||\; \Delta C{\color{ForestGreen}{B}}{\color{ForestGreen}{D}}$ , we can see that ${\color{ForestGreen}{DA}}$ matches with ${\color{ForestGreen}{BD}}$ .

So we have

$\begin{aligned} \frac{D A}{B D} & = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{DA}{BD} & = \dfrac{\square}{\square} \end{aligned}$

We were also given information about $CD$ and $CB$ . Since $\Delta {\color{red}{C}}{\color{red}{D}}A \;|||\; \Delta {\color{red}{C}}{\color{red}{B}}D$ , we can see that ${\color{red}{CD}}$ matches with ${\color{red}{CB}}$ .

So,

$\begin{aligned} \frac{D A}{B D} & = \frac{C D}{C B} (Δ C D A | | | Δ C B D) \end{aligned}$ $\begin{aligned} \dfrac{DA}{BD} & = \dfrac{CD}{CB} \;\; (\Delta CDA \;|||\; \Delta CBD) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Substitute in the lengths of the sides and solve the equation
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} ∴ \frac{D A}{12} & = \frac{13}{5} \\ ∴ D A & = 31,2 units \end{aligned}$ $\begin{aligned} \therefore {\color{ForestGreen}{\dfrac{DA}{12}}} & = {\color{red}{\dfrac{13}{5}}} \\ \therefore DA & = \text{31,2}\ \text{units} \end{aligned}$

Submit your answer as: and

Given similar triangles, calculate a side

In the diagram below, $\Delta SPQ \;|||\; \Delta SRP$ . Also, $SP = 10$ and $SR = 6$ .

Delphino has been asked to determine the length of $SQ$ . He knows that in order to calculate $SQ$ , he first needs to match up the sides of the two similar triangles.

Help Delphino to match up the sides in the similar triangles.

Answer:

Side in $\Delta SPQ$	Matching side in $\Delta SRP$
$SP$
$SQ$
$PQ$

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the similarity statement to choose the correct pairs of sides.

STEP: Use the similarity statement to match the sides
[−2 points ⇒ 0 / 2 points left]

Since $\Delta {\color{red}{S}}{\color{red}{P}}Q \;|||\; \Delta {\color{red}{S}}{\color{red}{R}}P$ , we can see that ${\color{red}{SP}}$ matches with ${\color{red}{SR}}$ . These are the first two letters in the similarity statement.

Also using $\Delta {\color{blue}{S}}P{\color{blue}{Q}} \;|||\; \Delta {\color{blue}{S}}R{\color{blue}{P}}$ , we can see that ${\color{blue}{SQ}}$ matches with ${\color{blue}{SP}}$ . These are the first and last letters in the similarity statement.

Finally, we can use $\Delta S{\color{ForestGreen}{P}}{\color{ForestGreen}{Q}} \;|||\; \Delta S{\color{ForestGreen}{R}}{\color{ForestGreen}{P}}$ to work out that ${\color{ForestGreen}{PQ}}$ matches with ${\color{ForestGreen}{RP}}$ . These are the second and third letters in the similarity statement.

Visually we can see the matching sides of the triangles:

Submit your answer as: andand

Determine the value of $SQ$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer: $SQ =$ $\text{units}$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similaity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, then their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $SQ$ . So, start by writing $SQ$ in the numerator:

$\begin{aligned} \frac{S Q}{◻} & = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{SQ}{\square} & = \dfrac{\square}{\square} \end{aligned}$

Since $\Delta {\color{blue}{S}}P{\color{blue}{Q}} \;|||\; \Delta {\color{blue}{S}}R{\color{blue}{P}}$ , we can see that ${\color{blue}{SQ}}$ matches with ${\color{blue}{SP}}$ .

So we have

$\begin{aligned} \frac{S Q}{S P} & = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{SQ}{SP} & = \dfrac{\square}{\square} \end{aligned}$

We were also given information about $SP$ and $SR$ . Since $\Delta {\color{red}{S}}{\color{red}{P}}Q \;|||\; \Delta {\color{red}{S}}{\color{red}{R}}P$ , we can see that ${\color{red}{SP}}$ matches with ${\color{red}{SR}}$ .

So,

$\begin{aligned} \frac{S Q}{S P} & = \frac{S P}{S R} (Δ S P Q | | | Δ S R P) \end{aligned}$ $\begin{aligned} \dfrac{SQ}{SP} & = \dfrac{SP}{SR} \;\; (\Delta SPQ \;|||\; \Delta SRP) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Substitute in the lengths of the sides and solve the equation
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} ∴ \frac{S Q}{10} & = \frac{10}{6} \\ ∴ S Q & = 16,66666... units \\ \approx 16,67 units \end{aligned}$ $\begin{aligned} \therefore {\color{blue}{\dfrac{SQ}{10}}} & = {\color{red}{\dfrac{10}{6}}} \\ \therefore SQ & = \text{16,66666...}\ \text{units} \\ & \approx \text{16,67}\ \text{units} \end{aligned}$

Submit your answer as: and

Given similar triangles, calculate a side

In the diagram below, $\Delta DAC \;|||\; \Delta BAD$ . Also, $DA = 12$ and $BA = 5$ .

Adeboye has been asked to determine the length of $AC$ . He knows that in order to calculate $AC$ , he first needs to match up the sides of the two similar triangles.

Help Adeboye to match up the sides in the similar triangles.

Answer:

Side in $\Delta DAC$	Matching side in $\Delta BAD$
$DA$
$DC$
$AC$

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Use the similarity statement to choose the correct pairs of sides.

STEP: Use the similarity statement to match the sides
[−2 points ⇒ 0 / 2 points left]

Since $\Delta {\color{red}{D}}{\color{red}{A}}C \;|||\; \Delta {\color{red}{B}}{\color{red}{A}}D$ , we can see that ${\color{red}{DA}}$ matches with ${\color{red}{BA}}$ . These are the first two letters in the similarity statement.

Also using $\Delta {\color{blue}{D}}A{\color{blue}{C}} \;|||\; \Delta {\color{blue}{B}}A{\color{blue}{D}}$ , we can see that ${\color{blue}{DC}}$ matches with ${\color{blue}{BD}}$ . These are the first and last letters in the similarity statement.

Finally, we can use $\Delta D{\color{ForestGreen}{A}}{\color{ForestGreen}{C}} \;|||\; \Delta B{\color{ForestGreen}{A}}{\color{ForestGreen}{D}}$ to work out that ${\color{ForestGreen}{AC}}$ matches with ${\color{ForestGreen}{AD}}$ . These are the second and third letters in the similarity statement.

Visually we can see the matching sides of the triangles:

Submit your answer as: andand

Determine the value of $AC$ .

The diagram is repeated here for your convenience.

INSTRUCTION: Round your answer to two decimal places, if necessary.

Answer: $AC =$ $\text{units}$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similaity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, then their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $AC$ . So, start by writing $AC$ in the numerator:

$\begin{aligned} \frac{A C}{◻} & = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{AC}{\square} & = \dfrac{\square}{\square} \end{aligned}$

Since $\Delta D{\color{ForestGreen}{A}}{\color{ForestGreen}{C}} \;|||\; \Delta B{\color{ForestGreen}{A}}{\color{ForestGreen}{D}}$ , we can see that ${\color{ForestGreen}{AC}}$ matches with ${\color{ForestGreen}{AD}}$ .

So we have

$\begin{aligned} \frac{A C}{A D} & = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{AC}{AD} & = \dfrac{\square}{\square} \end{aligned}$

We were also given information about $DA$ and $BA$ . Since $\Delta {\color{red}{D}}{\color{red}{A}}C \;|||\; \Delta {\color{red}{B}}{\color{red}{A}}D$ , we can see that ${\color{red}{DA}}$ matches with ${\color{red}{BA}}$ .

So,

$\begin{aligned} \frac{A C}{A D} & = \frac{D A}{B A} (Δ D A C | | | Δ B A D) \end{aligned}$ $\begin{aligned} \dfrac{AC}{AD} & = \dfrac{DA}{BA} \;\; (\Delta DAC \;|||\; \Delta BAD) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So, we write this as the reason.

STEP: Substitute in the lengths of the sides and solve the equation
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} ∴ \frac{A C}{12} & = \frac{12}{5} \\ ∴ A C & = 28,8 units \end{aligned}$ $\begin{aligned} \therefore {\color{ForestGreen}{\dfrac{AC}{12}}} & = {\color{red}{\dfrac{12}{5}}} \\ \therefore AC & = \text{28,8}\ \text{units} \end{aligned}$

Submit your answer as: and

Similarity and congruency notation

Consider the following triangles:

Answer:

Which of the following best describes the relationship between $\Delta DEF$ and $\Delta JKL$ ?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The symbol $\; \equiv \;$ is used to indicate that two triangles are congruent. The symbol $\;|||\;$ is used to indicate that two triangles are similar.

STEP: Identify the triangle relationship
[−2 points ⇒ 0 / 2 points left]

Two triangles are similar if they are the same shape, but they can be a different size. The symbol for similarity is $|||$ . There are two ways to know if triangles are similar:

if their sides are in proportion. This means that the sides in one triangle have been multiplied by the same number to get the sides in another triangle.
if all of their matching angles are equal.

Two triangles are congruent if they are exactly the same size and shape. The symbol for congruency is $\equiv$ . There are four ways to know if triangles are congruent: SSS, SAS, SAA, or $90°$ HS.

TIP: When two triangles are equal in all respects, we say that they are congruent and use the symbol for congruency (

\equiv

$\equiv$ ). We never use the equals sign (

=

$=$ ) to compare two triangles.

We can see that the sides in $\Delta DEF$ have been multiplied by $\text{2}$ to get the matching sides in $\Delta JKL$ .

So, the triangles are similar, and we write:

$\Delta DEF \;|||\; \Delta JKL$ (sides of $\Delta$ in prop).

Submit your answer as:

Δ D E F | | | Δ J K L

$\Delta DEF \;|||\; \Delta JKL$

Similarity and congruency notation

Consider the following triangles:

Answer:

Which of the following best describes the relationship between $\Delta DEF$ and $\Delta JKL$ ?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The symbol $\; \equiv \;$ is used to indicate that two triangles are congruent. The symbol $\;|||\;$ is used to indicate that two triangles are similar.

STEP: Identify the triangle relationship
[−2 points ⇒ 0 / 2 points left]

Two triangles are similar if they are the same shape, but they can be a different size. The symbol for similarity is $|||$ . There are two ways to know if triangles are similar:

if their sides are in proportion. This means that the sides in one triangle have been multiplied by the same number to get the sides in another triangle.
if all of their matching angles are equal.

Two triangles are congruent if they are exactly the same size and shape. The symbol for congruency is $\equiv$ . There are four ways to know if triangles are congruent: SSS, SAS, SAA, or $90°$ HS.

TIP: When two triangles are equal in all respects, we say that they are congruent and use the symbol for congruency (

\equiv

$\equiv$ ). We never use the equals sign (

=

$=$ ) to compare two triangles.

We were given one matching side and two matching angles in both triangles. So, the triangles are congruent, and we write:

$\Delta DEF \equiv \Delta JKL$ (SAA)

NOTE: If two triangles are congruent, then they are always also similar. But, being congruent is more special that being similar, so the best way to describe them is to say that they are congruent.

Submit your answer as:

Δ D E F \equiv Δ J K L

$\Delta DEF \equiv \Delta JKL$

Similarity and congruency notation

Consider the following triangles:

Answer:

Which of the following best describes the relationship between $\Delta DEF$ and $\Delta JKL$ ?

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

The symbol $\; \equiv \;$ is used to indicate that two triangles are congruent. The symbol $\;|||\;$ is used to indicate that two triangles are similar.

STEP: Identify the triangle relationship
[−2 points ⇒ 0 / 2 points left]

Two triangles are similar if they are the same shape, but they can be a different size. The symbol for similarity is $|||$ . There are two ways to know if triangles are similar:

if their sides are in proportion. This means that the sides in one triangle have been multiplied by the same number to get the sides in another triangle.
if all of their matching angles are equal.

Two triangles are congruent if they are exactly the same size and shape. The symbol for congruency is $\equiv$ . There are four ways to know if triangles are congruent: SSS, SAS, SAA, or $90°$ HS.

TIP: When two triangles are equal in all respects, we say that they are congruent and use the symbol for congruency (

\equiv

$\equiv$ ). We never use the equals sign (

=

$=$ ) to compare two triangles.

We were given two sides and the included angle in both triangles. So, the triangles are congruent, and we write:

$\Delta DEF \equiv \Delta JKL$ (SAS)

NOTE: If two triangles are congruent, then they are always also similar. But, being congruent is more special that being similar, so the best way to describe them is to say that they are congruent.

Submit your answer as:

Δ D E F \equiv Δ J K L

$\Delta DEF \equiv \Delta JKL$

Determine the proportionality factor between similar triangles

The diagram below shows two similar triangles:

All of the sides in $\Delta KLJ$ been multiplied by the same number, $k$ , to get the sides in $\Delta PMN$ . What is the value of $k$ ?

INSTRUCTION: Give your answer as an exact decimal or as a fraction.

Answer:

k =

$k =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Divide the longest side in $\Delta PMN$ by the longest side in $\Delta KLJ$ .

STEP: Divide the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ K L J

$\Delta KLJ$
[−2 points ⇒ 0 / 2 points left]

We are looking for the number that each matching side in $\Delta KLJ$ has been multiplied by, to get the matching side in $\Delta PMN$ .

NOTE: We call this number the proportionality constant,

k

$k$ .

The longest side in $\Delta KLJ$ is ${\color{blue}{24}}$ . It must match up with the longest side in $\Delta PMN$ , which is ${\color{blue}{\text{30}}}$ . We will use these sides to determine the multiplier. Let the proportionality constant be $k$ :

\begin{aligned} 24 \times k & = 30 \\ ∴ \frac{24 \times k}{24} & = \frac{30}{24} \\ ∴ k & = 1,25 \end{aligned}

$\begin{aligned} {\color{blue}{24}} \times k & = {\color{blue}{\text{30}}} \\[0.2cm] \therefore \dfrac{24 \times k}{24} & = \dfrac{\text{30}}{24} \\[0.2cm] \therefore k & = \text{1,25} \end{aligned}$

NOTE: In the last step, we divided the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ K L J

$\Delta KLJ$ , to find

k

$k$ .

We can check that this value works for the other sides:

The shortest sides, ${\color{red}{KL}}$ and ${\color{red}{PM}}$ : $\;\;{\color{red}{12}} \times \text{1,25} = {\color{red}{\text{15}}}$
The middle sides, ${\color{ForestGreen}{LJ}}$ and ${\color{ForestGreen}{MN}}$ : $\;\;{\color{ForestGreen}{20}} \times \text{1,25} = {\color{ForestGreen}{\text{25}}}$

So, all the sides in $\Delta KLJ$ have been multiplied by $\text{1,25}$ to get the sides in $\Delta PMN$ . We call this number the proportionality constant.

Submit your answer as:

Determine the proportionality factor between similar triangles

The diagram below shows two similar triangles:

All of the sides in $\Delta CBA$ been multiplied by the same number, $k$ , to get the sides in $\Delta DFE$ . What is the value of $k$ ?

INSTRUCTION: Give your answer as an exact decimal or as a fraction.

Answer:

k =

$k =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Divide the longest side in $\Delta DFE$ by the longest side in $\Delta CBA$ .

STEP: Divide the longest side in

Δ D F E

$\Delta DFE$ by the longest side in

Δ C B A

$\Delta CBA$
[−2 points ⇒ 0 / 2 points left]

We are looking for the number that each matching side in $\Delta CBA$ has been multiplied by, to get the matching side in $\Delta DFE$ .

NOTE: We call this number the proportionality constant,

k

$k$ .

The longest side in $\Delta CBA$ is ${\color{blue}{14}}$ . It must match up with the longest side in $\Delta DFE$ , which is ${\color{blue}{\text{21}}}$ . We will use these sides to determine the multiplier. Let the proportionality constant be $k$ :

\begin{aligned} 14 \times k & = 21 \\ ∴ \frac{14 \times k}{14} & = \frac{21}{14} \\ ∴ k & = 1,5 \end{aligned}

$\begin{aligned} {\color{blue}{14}} \times k & = {\color{blue}{\text{21}}} \\[0.2cm] \therefore \dfrac{14 \times k}{14} & = \dfrac{\text{21}}{14} \\[0.2cm] \therefore k & = \text{1,5} \end{aligned}$

NOTE: In the last step, we divided the longest side in

Δ D F E

$\Delta DFE$ by the longest side in

Δ C B A

$\Delta CBA$ , to find

k

$k$ .

We can check that this value works for the other sides:

The shortest sides, ${\color{red}{CB}}$ and ${\color{red}{DF}}$ : $\;\;{\color{red}{6}} \times \text{1,5} = {\color{red}{\text{9}}}$
The middle sides, ${\color{ForestGreen}{BA}}$ and ${\color{ForestGreen}{FE}}$ : $\;\;{\color{ForestGreen}{12}} \times \text{1,5} = {\color{ForestGreen}{\text{18}}}$

So, all the sides in $\Delta CBA$ have been multiplied by $\text{1,5}$ to get the sides in $\Delta DFE$ . We call this number the proportionality constant.

Submit your answer as:

Determine the proportionality factor between similar triangles

The diagram below shows two similar triangles:

All of the sides in $\Delta KLJ$ been multiplied by the same number, $k$ , to get the sides in $\Delta PMN$ . What is the value of $k$ ?

INSTRUCTION: Give your answer as an exact decimal or as a fraction.

Answer:

k =

$k =$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Divide the longest side in $\Delta PMN$ by the longest side in $\Delta KLJ$ .

STEP: Divide the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ K L J

$\Delta KLJ$
[−2 points ⇒ 0 / 2 points left]

We are looking for the number that each matching side in $\Delta KLJ$ has been multiplied by, to get the matching side in $\Delta PMN$ .

NOTE: We call this number the proportionality constant,

k

$k$ .

The longest side in $\Delta KLJ$ is ${\color{blue}{15}}$ . It must match up with the longest side in $\Delta PMN$ , which is ${\color{blue}{\text{22,5}}}$ . We will use these sides to determine the multiplier. Let the proportionality constant be $k$ :

\begin{aligned} 15 \times k & = 22,5 \\ ∴ \frac{15 \times k}{15} & = \frac{22,5}{15} \\ ∴ k & = 1,5 \end{aligned}

$\begin{aligned} {\color{blue}{15}} \times k & = {\color{blue}{\text{22,5}}} \\[0.2cm] \therefore \dfrac{15 \times k}{15} & = \dfrac{\text{22,5}}{15} \\[0.2cm] \therefore k & = \text{1,5} \end{aligned}$

NOTE: In the last step, we divided the longest side in

Δ P M N

$\Delta PMN$ by the longest side in

Δ K L J

$\Delta KLJ$ , to find

k

$k$ .

We can check that this value works for the other sides:

The shortest sides, ${\color{red}{KL}}$ and ${\color{red}{PM}}$ : $\;\;{\color{red}{9}} \times \text{1,5} = {\color{red}{\text{13,5}}}$
The middle sides, ${\color{ForestGreen}{LJ}}$ and ${\color{ForestGreen}{MN}}$ : $\;\;{\color{ForestGreen}{14}} \times \text{1,5} = {\color{ForestGreen}{\text{21}}}$

So, all the sides in $\Delta KLJ$ have been multiplied by $\text{1,5}$ to get the sides in $\Delta PMN$ . We call this number the proportionality constant.

Submit your answer as:

Similar, congruent, or neither?

Consider the following polygons:

Are the polygons similar, congruent, or neither? Choose the most correct description.

Answer:

The two polygons are:

similar
congruent
not similar or congruent

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.
Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Identify if the polygons are similar, congruent, or neither
[−1 point ⇒ 0 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal in size and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal.

The two polygons do not have to be in the same orientation. This means that they can face different directions.

The matching sides and angles in the two polygons are exactly equal.

So, the two polygons are congruent.

NOTE: Congruent polygons are always also similar to each other. But if two polygons are exactly equal, it is more correct to call them congruent.

Submit your answer as:

congruent

Similar, congruent, or neither?

Consider the following polygons:

Are the polygons similar, congruent, or neither? Choose the most correct description.

Answer:

The two polygons are:

similar
congruent
not similar or congruent

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.
Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Identify if the polygons are similar, congruent, or neither
[−1 point ⇒ 0 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal in size and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal.

The two polygons do not have to be in the same orientation. This means that they can face different directions.

The matching sides and angles in the two polygons are exactly equal.

So, the two polygons are congruent.

NOTE: Congruent polygons are always also similar to each other. But if two polygons are exactly equal, it is more correct to call them congruent.

Submit your answer as:

congruent

Similar, congruent, or neither?

Consider the following polygons:

Are the polygons similar, congruent, or neither? Choose the most correct description.

Answer:

The two polygons are:

similar
congruent
not similar or congruent

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal and all of their matching sides are in proportion.
Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Identify if the polygons are similar, congruent, or neither
[−1 point ⇒ 0 / 1 points left]

Two polygons are similar if they are the same shape, but they may be different sizes. All of their matching angles are equal in size and all of their matching sides are in proportion.

Two polygons are congruent if they are exactly the same size and shape. All their matching sides and angles are equal.

The two polygons do not have to be in the same orientation. This means that they can face different directions.

The matching angles in the two quadrilaterals are exactly the same. We can also see that all of the sides in the first quadrilateral have been multiplied by $\text{0,5}$ to get the sides in the second one. This means that the sides are in proportion.

This is easier to see if the quadrilaterals are in the same orientation.

So, the two polygons are similar.

Submit your answer as:

similar

Identify similar triangles

$\Delta CAB$ is drawn below.

Which of the following triangles is definitely similar to $\Delta CAB$ ? Give a reason for the similarity.

Answer:

$\Delta CAB \;|||\; \Delta$

string

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Triangles are similar if either their:

matching sides are in proportion (sides of $\Delta$ in prop), or
matching angles are equal (equiangular $\Delta$ s).

STEP: Identify the similar triangles
[−1 point ⇒ 1 / 2 points left]

Triangles are similar if either their:

sides are in proportion (sides of $\Delta$ in prop), or
angles are equal (equiangular $\Delta$ s).

If we divide the matching sides in $\Delta YXZ$ by the matching sides in $\Delta CAB$ , we get the same number each time:

\begin{aligned} longest sides: & \frac{Y Z}{C B} & = \frac{14}{7} & = 2 \\ middle sides: & \frac{X Z}{A B} & = \frac{10}{5} & = 2 \\ shortest sides: & \frac{Y X}{C A} & = \frac{6}{3} & = 2 \end{aligned}

$\begin{alignedat}{3} {\color{red}{\text{longest sides: }}} \;\; & \dfrac{YZ}{CB} & = \dfrac{\text{14}}{7} & = \text{2} \\[0.2cm] {\color{blue}{\text{middle sides: }}} \;\; & \dfrac{XZ}{AB} & = \dfrac{\text{10}}{5} & = \text{2} \\[0.2cm] {\color{ForestGreen}{\text{shortest sides: }}} \;\; & \dfrac{YX}{CA} & = \dfrac{\text{6}}{3} & = \text{2} \end{alignedat}$

So, these triangles are similar, because their sides are in proportion.

STEP: Match up the triangle vertices
[−1 point ⇒ 0 / 2 points left]

We must name the triangles in the order that their sides match up.

In $\Delta CAB$ , Point $C$ is between the longest and the shortest sides. In the same way, in $\Delta YXZ$ Point $Y$ is between the longest and the shortest sides.

So, Point $C$ matches with Point $Y$ .

In $\Delta CAB$ , Point $A$ is the other end of the shortest side. In the same way, in $\Delta YXZ$ Point $X$ is the other end of the shortest side. So, Point $A$ matches with Point $X$ .

Then, the last vertices must also match up. So, Point $B$ matches with Point $Z$ , and we write:

$\Delta CAB \;|||\; \Delta YXZ$ (sides of $\Delta$ in prop).

Submit your answer as: and

Identify similar triangles

$\Delta PQR$ is drawn below.

Which of the following triangles is definitely similar to $\Delta PQR$ ? Give a reason for the similarity.

Answer:

$\Delta PQR \;|||\; \Delta$

string

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Triangles are similar if either their:

matching sides are in proportion (sides of $\Delta$ in prop), or
matching angles are equal (equiangular $\Delta$ s).

STEP: Identify the similar triangles
[−1 point ⇒ 1 / 2 points left]

Triangles are similar if either their:

sides are in proportion (sides of $\Delta$ in prop), or
angles are equal (equiangular $\Delta$ s).

We were given three pairs of equal angles in the two triangles, so they are similar.

STEP: Match up the triangle vertices
[−1 point ⇒ 0 / 2 points left]

We must name the triangles in the order that their angles match up.

We can see that the angle at ${\color{red}{P}}$ is equal to the angle at ${\color{red}{Y}}$ . So, Point $P$ matches with Point $Y$ .

In the same way:

Point ${\color{blue}{Q}}$ matches with ${\color{blue}{Z}}$ , and
Point ${\color{ForestGreen}{R}}$ matches with ${\color{ForestGreen}{X}}$ .

So,

$\Delta PQR \;|||\; \Delta YZX$ (equiangular $\Delta$ s)

NOTE: Equiangular comes from the equal angles in both triangles. It looks similar to equilateral, which means equal lines. But, the two words do not mean the same thing.

Submit your answer as: and

Identify similar triangles

$\Delta RPQ$ is drawn below.

Which of the following triangles is definitely similar to $\Delta RPQ$ ? Give a reason for the similarity.

Answer:

$\Delta RPQ \;|||\; \Delta$

string

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Triangles are similar if either their:

matching sides are in proportion (sides of $\Delta$ in prop), or
matching angles are equal (equiangular $\Delta$ s).

STEP: Identify the similar triangles
[−1 point ⇒ 1 / 2 points left]

Triangles are similar if either their:

sides are in proportion (sides of $\Delta$ in prop), or
angles are equal (equiangular $\Delta$ s).

We were given three pairs of equal angles in the two triangles, so they are similar.

STEP: Match up the triangle vertices
[−1 point ⇒ 0 / 2 points left]

We must name the triangles in the order that their angles match up.

We can see that the angle at ${\color{red}{R}}$ is equal to the angle at ${\color{red}{A}}$ . So, Point $R$ matches with Point $A$ .

In the same way:

Point ${\color{blue}{P}}$ matches with ${\color{blue}{B}}$ , and
Point ${\color{ForestGreen}{Q}}$ matches with ${\color{ForestGreen}{C}}$ .

So,

$\Delta RPQ \;|||\; \Delta ABC$ (equiangular $\Delta$ s)

NOTE: Equiangular comes from the equal angles in both triangles. It looks similar to equilateral, which means equal lines. But, the two words do not mean the same thing.

Submit your answer as: and

Prove simple similarity with sides

Consider the following triangles:

Prove that $\Delta PQR \;|||\; \Delta ZYX$ .

INSTRUCTION: Give your answer as a simplified fraction or decimal.

Answer:

In $\Delta PQR$ and $\Delta ZYX$ :

$ZY \div$ $=$
$YX \div$ $=$
$ZX \div$ $=$

$\therefore \Delta PQR \;|||\; \Delta ZYX$ (sides of $\Delta$ in prop)

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

Which of these strategies is more appropriate for this question?

STEP: Match up sides and find the proportionality constant
[−3 points ⇒ 1 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

In this question, we have more information about the sides of the triangles than about the angles. So, we will prove that all of the sides are in proportion.

We must show that each side must be multiplied by the same number to make its matching side in the other triangle.

Firstly, we must match up the sides:

The longest sides ( ${\color{red}{ZY}}$ and ${\color{red}{PQ}}$ ) match up.
The shortest sides ( ${\color{blue}{ZX}}$ and ${\color{blue}{PR}}$ ) match up.
${\color{ForestGreen}{XY}}$ and ${\color{ForestGreen}{QR}}$ must match up because they are the only sides left.

Now we will work out what each side in $\Delta PQR$ should be multiplied by:

\begin{aligned} \frac{Z Y}{P Q} & = \frac{28}{16} = 1,75 \\ \frac{Y X}{Q R} & = \frac{21}{12} = 1,75 \\ \frac{Z X}{P R} & = \frac{14}{8} = 1,75 \end{aligned}

$\begin{aligned} \dfrac{ZY}{PQ} & = \dfrac{\text{28}}{\text{16}} = \text{1,75} \\[0.2cm] \dfrac{YX}{QR} & = \dfrac{\text{21}}{\text{12}} = \text{1,75} \\[0.2cm] \dfrac{ZX}{PR} & = \dfrac{\text{14}}{\text{8}} = \text{1,75} \end{aligned}$

This means that any side in $\Delta PQR$ must be multiplied by $\text{1,75}$ to make its matching side in $\Delta ZYX$ .

For example, $\text{16} \times \text{1,75} = \text{28}$ .

We call this number the proportionality constant.

If any one of these division sums had given us a different answer, the sides of the triangles would not have been in proportion. Then, the triangles would not be similar.

TIP: If your calculations tell you that the triangles are not similar in a question where you are asked to prove similarity, you probably made a mistake somewhere. Check your working out!

STEP: Prove similarity
[−1 point ⇒ 0 / 4 points left]

Now that we have matched up the sides, and worked out the proportionality constant, we just need to put the information into a formal similarity proof:

In $\Delta PQR$ and $\Delta ZYX$ :

$ZY \div PQ$ $=$ $\dfrac{\text{28}}{\text{16}}$ $=$ $\text{1,75}$
$YX \div QR$ $=$ $\dfrac{\text{21}}{\text{12}}$ $=$ $\text{1,75}$
$ZX \div PR$ $=$ $\dfrac{\text{14}}{\text{8}}$ $=$ $\text{1,75}$

$\therefore \Delta PQR \;|||\; \Delta ZYX$ (sides of $\Delta$ in prop)

Submit your answer as: andandandandand

Prove simple similarity with sides

Consider the following triangles:

Prove that $\Delta BAC \;|||\; \Delta EDF$ .

INSTRUCTION: Give your answer as a simplified fraction or decimal.

Answer:

In $\Delta BAC$ and $\Delta EDF$ :

$ED \div$ $=$
$DF \div$ $=$
$EF \div$ $=$

$\therefore \Delta BAC \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

Which of these strategies is more appropriate for this question?

STEP: Match up sides and find the proportionality constant
[−3 points ⇒ 1 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

In this question, we have more information about the sides of the triangles than about the angles. So, we will prove that all of the sides are in proportion.

We must show that each side must be multiplied by the same number to make its matching side in the other triangle.

Firstly, we must match up the sides:

The longest sides ( ${\color{red}{ED}}$ and ${\color{red}{BA}}$ ) match up.
The shortest sides ( ${\color{blue}{EF}}$ and ${\color{blue}{BC}}$ ) match up.
${\color{ForestGreen}{FD}}$ and ${\color{ForestGreen}{AC}}$ must match up because they are the only sides left.

Now we will work out what each side in $\Delta BAC$ should be multiplied by:

\begin{aligned} \frac{E D}{B A} & = \frac{54}{24} = 2,25 \\ \frac{D F}{A C} & = \frac{45}{20} = 2,25 \\ \frac{E F}{B C} & = \frac{27}{12} = 2,25 \end{aligned}

$\begin{aligned} \dfrac{ED}{BA} & = \dfrac{\text{54}}{\text{24}} = \text{2,25} \\[0.2cm] \dfrac{DF}{AC} & = \dfrac{\text{45}}{\text{20}} = \text{2,25} \\[0.2cm] \dfrac{EF}{BC} & = \dfrac{\text{27}}{\text{12}} = \text{2,25} \end{aligned}$

This means that any side in $\Delta BAC$ must be multiplied by $\text{2,25}$ to make its matching side in $\Delta EDF$ .

For example, $\text{24} \times \text{2,25} = \text{54}$ .

We call this number the proportionality constant.

If any one of these division sums had given us a different answer, the sides of the triangles would not have been in proportion. Then, the triangles would not be similar.

TIP: If your calculations tell you that the triangles are not similar in a question where you are asked to prove similarity, you probably made a mistake somewhere. Check your working out!

STEP: Prove similarity
[−1 point ⇒ 0 / 4 points left]

Now that we have matched up the sides, and worked out the proportionality constant, we just need to put the information into a formal similarity proof:

In $\Delta BAC$ and $\Delta EDF$ :

$ED \div BA$ $=$ $\dfrac{\text{54}}{\text{24}}$ $=$ $\text{2,25}$
$DF \div AC$ $=$ $\dfrac{\text{45}}{\text{20}}$ $=$ $\text{2,25}$
$EF \div BC$ $=$ $\dfrac{\text{27}}{\text{12}}$ $=$ $\text{2,25}$

$\therefore \Delta BAC \;|||\; \Delta EDF$ (sides of $\Delta$ in prop)

Submit your answer as: andandandandand

Prove simple similarity with sides

Consider the following triangles:

Prove that $\Delta JKL \;|||\; \Delta NMP$ .

INSTRUCTION: Give your answer as a simplified fraction or decimal.

Answer:

In $\Delta JKL$ and $\Delta NMP$ :

$NM \div$ $=$
$MP \div$ $=$
$NP \div$ $=$

$\therefore \Delta JKL \;|||\; \Delta NMP$ (sides of $\Delta$ in prop)

numeric

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

Which of these strategies is more appropriate for this question?

STEP: Match up sides and find the proportionality constant
[−3 points ⇒ 1 / 4 points left]

There are two ways of proving that triangles are similar:

Prove that they are equiangular.
Prove that their sides are in proportion.

In this question, we have more information about the sides of the triangles than about the angles. So, we will prove that all of the sides are in proportion.

We must show that each side must be multiplied by the same number to make its matching side in the other triangle.

Firstly, we must match up the sides:

The longest sides ( ${\color{red}{NM}}$ and ${\color{red}{JK}}$ ) match up.
The shortest sides ( ${\color{blue}{NP}}$ and ${\color{blue}{JL}}$ ) match up.
${\color{ForestGreen}{PM}}$ and ${\color{ForestGreen}{KL}}$ must match up because they are the only sides left.

Now we will work out what each side in $\Delta JKL$ should be multiplied by:

\begin{aligned} \frac{N M}{J K} & = \frac{35}{14} = 2,5 \\ \frac{M P}{K L} & = \frac{27,5}{11} = 2,5 \\ \frac{N P}{J L} & = \frac{20}{8} = 2,5 \end{aligned}

$\begin{aligned} \dfrac{NM}{JK} & = \dfrac{\text{35}}{\text{14}} = \text{2,5} \\[0.2cm] \dfrac{MP}{KL} & = \dfrac{\text{27,5}}{\text{11}} = \text{2,5} \\[0.2cm] \dfrac{NP}{JL} & = \dfrac{\text{20}}{\text{8}} = \text{2,5} \end{aligned}$

This means that any side in $\Delta JKL$ must be multiplied by $\text{2,5}$ to make its matching side in $\Delta NMP$ .

For example, $\text{14} \times \text{2,5} = \text{35}$ .

We call this number the proportionality constant.

If any one of these division sums had given us a different answer, the sides of the triangles would not have been in proportion. Then, the triangles would not be similar.

TIP: If your calculations tell you that the triangles are not similar in a question where you are asked to prove similarity, you probably made a mistake somewhere. Check your working out!

STEP: Prove similarity
[−1 point ⇒ 0 / 4 points left]

Now that we have matched up the sides, and worked out the proportionality constant, we just need to put the information into a formal similarity proof:

In $\Delta JKL$ and $\Delta NMP$ :

$NM \div JK$ $=$ $\dfrac{\text{35}}{\text{14}}$ $=$ $\text{2,5}$
$MP \div KL$ $=$ $\dfrac{\text{27,5}}{\text{11}}$ $=$ $\text{2,5}$
$NP \div JL$ $=$ $\dfrac{\text{20}}{\text{8}}$ $=$ $\text{2,5}$

$\therefore \Delta JKL \;|||\; \Delta NMP$ (sides of $\Delta$ in prop)

Submit your answer as: andandandandand

Basic deductions from similar triangles

In the diagram below, $\Delta RQP \;|||\; \Delta DEF$ , because their sides are in proportion.

What can you work out about the angles in the two triangles?
Answer:
- $\angle P$ $\; = \angle$ $\;\;(\Delta RQP \;|||\; \Delta DEF)$
- $\angle Q$ $\; = \angle$ $\;\;(\Delta RQP \;|||\; \Delta DEF)$
- $\angle R$ $\; = \angle$ $\;\;(\Delta RQP \;|||\; \Delta DEF)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the sides of two triangles are in proportion, then their matching angles are equal. Can you match up the angles of the two triangles?

STEP: Match up the angles
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta RQP$ is similar to $\Delta DEF$ , because the sides are in proportion. We can see that the all of the sides in $\Delta RQP$ have been multiplied by $\text{2}$ to get the sides in $\Delta DEF$ .

If the sides of two triangles are in proportion, then all of their matching angles will be equal. We can identify the matching angles using the similarity statement: $\Delta RQP \;|||\; \Delta DEF$ .

So,
- ${\color{blue}{\hat{R}}} = {\color{blue}{\hat{D}}} \;\;(\Delta {\color{blue}{R}}QP \;|||\; \Delta {\color{blue}{D}}EF)$
- ${\color{red}{\hat{Q}}} = {\color{red}{\hat{E}}} \;\;(\Delta R{\color{red}{Q}}P \;|||\; \Delta D{\color{red}{E}}F)$
- ${\color{ForestGreen}{\hat{P}}} = {\color{ForestGreen}{\hat{F}}} \;\;(\Delta RQ{\color{ForestGreen}{P}} \;|||\; \Delta DE{\color{ForestGreen}{F}})$
NOTE: We only know that the angles are equal because the triangles are similar. So, we write this as the reason next to the deduction.

Submit your answer as: andand
In the diagram below, $\Delta CBA \;|||\; \Delta MNL$ , because their angles are equal.

What can you deduce about the sides of the two triangles? Choose the correct option.
Answer:
$\dfrac{MN}{CB} = \dfrac{BA}{NL} = \dfrac{CA}{ML}$ $\;\;(\Delta CBA \;|||\; \Delta MNL)$

$\dfrac{CB}{MN} = \dfrac{BA}{NL} = \dfrac{CA}{ML}$ $\;\;(\Delta CBA \;|||\; \Delta MNL)$

$\dfrac{\hat{C}}{\hat{M}} = \dfrac{\hat{B}}{\hat{N}} = \dfrac{\hat{A}}{\hat{L}}$ $\;\;(\Delta CBA \;|||\; \Delta MNL)$

$\dfrac{BA}{MN} = \dfrac{CB}{NL} = \dfrac{CA}{ML}$ $\;\;(\Delta CBA \;|||\; \Delta MNL)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the angles in two triangles are equal, then their sides will be in proportion. Which of the options shows the correct ratios?

STEP: Match up the sides
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta CBA$ is similar to $\Delta MNL$ , because the matching angles are equal.

If the matching angles in two triangles are equal, then their sides are in proportion. So, when we divide each side of one triangle by its partner in the other triangle, we always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

Since $\Delta {\color{blue}{C}}{\color{blue}{B}}A \;|||\; \Delta {\color{blue}{M}}{\color{blue}{N}}L$ , we can see that ${\color{blue}{CB}}$ matches with ${\color{blue}{MN}}$ .

In the same way, we can see that ${\color{ForestGreen}{CA}}$ matches with ${\color{ForestGreen}{ML}}$ $(\Delta {\color{ForestGreen}{C}}B{\color{ForestGreen}{A}} \;|||\; \Delta {\color{ForestGreen}{M}}N{\color{ForestGreen}{L}})$ .

Finally, the last sides must also match up. So ${\color{red}{BA}}$ matches with ${\color{red}{NL}}$ $(\Delta C{\color{red}{B}}{\color{red}{A}} \;|||\; \Delta M{\color{red}{N}}{\color{red}{L}})$ .

TIP: We must always divide the triangle sides in the same order. So, if we put $\Delta CBA$ 's sides in the numerator for the first fraction, then $\Delta CBA$ 's sides must always be in the numerator.

We can also see the matching sides in the diagram:

So, the correct answer is

$\begin{matrix} \frac{C B}{M N} = \frac{B A}{N L} = \frac{C A}{M L} (Δ C B A | | | Δ M N L) \end{matrix}$ $\begin{array}{c} \dfrac{CB}{MN} = \dfrac{BA}{NL} = \dfrac{CA}{ML} \;\; (\Delta CBA \;|||\; \Delta MNL) \end{array}$

NOTE: We only know that the sides are in proportion because the triangles are similar. So, we write this as the reason.

There are lots of correct ways to write the ratios. These are two more correct examples:
1. $\dfrac{MN}{CB} = \dfrac{NL}{BA} = \dfrac{ML}{CA} \;\; (\Delta CBA \;|||\; \Delta MNL)$
2. $\dfrac{BA}{NL} = \dfrac{CA}{ML} = \dfrac{CB}{MN} \;\; (\Delta CBA \;|||\; \Delta MNL)$
Submit your answer as:
$\dfrac{CB}{MN} = \dfrac{BA}{NL} = \dfrac{CA}{ML}$ $\;\;(\Delta CBA \;|||\; \Delta MNL)$

Basic deductions from similar triangles

In the diagram below, $\Delta ACB \;|||\; \Delta MNL$ , because their sides are in proportion.

What can you work out about the angles in the two triangles?
Answer:
- $\angle B$ $\; = \angle$ $\;\;(\Delta ACB \;|||\; \Delta MNL)$
- $\angle A$ $\; = \angle$ $\;\;(\Delta ACB \;|||\; \Delta MNL)$
- $\angle C$ $\; = \angle$ $\;\;(\Delta ACB \;|||\; \Delta MNL)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the sides of two triangles are in proportion, then their matching angles are equal. Can you match up the angles of the two triangles?

STEP: Match up the angles
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta ACB$ is similar to $\Delta MNL$ , because the sides are in proportion. We can see that the all of the sides in $\Delta ACB$ have been multiplied by $\text{2}$ to get the sides in $\Delta MNL$ .

If the sides of two triangles are in proportion, then all of their matching angles will be equal. We can identify the matching angles using the similarity statement: $\Delta ACB \;|||\; \Delta MNL$ .

So,
- ${\color{blue}{\hat{A}}} = {\color{blue}{\hat{M}}} \;\;(\Delta {\color{blue}{A}}CB \;|||\; \Delta {\color{blue}{M}}NL)$
- ${\color{red}{\hat{C}}} = {\color{red}{\hat{N}}} \;\;(\Delta A{\color{red}{C}}B \;|||\; \Delta M{\color{red}{N}}L)$
- ${\color{ForestGreen}{\hat{B}}} = {\color{ForestGreen}{\hat{L}}} \;\;(\Delta AC{\color{ForestGreen}{B}} \;|||\; \Delta MN{\color{ForestGreen}{L}})$
NOTE: We only know that the angles are equal because the triangles are similar. So, we write this as the reason next to the deduction.

Submit your answer as: andand
In the diagram below, $\Delta PRQ \;|||\; \Delta DEF$ , because their angles are equal.

What can you deduce about the sides of the two triangles? Choose the correct option.
Answer:
$\dfrac{\hat{D}}{\hat{P}} = \dfrac{\hat{E}}{\hat{R}} = \dfrac{\hat{F}}{\hat{Q}}$ $\;\;(\Delta PRQ \;|||\; \Delta DEF)$

$\dfrac{EF}{RQ} = \dfrac{DF}{PQ} = \dfrac{DE}{PR}$ $\;\;(\Delta PRQ \;|||\; \Delta DEF)$

$\dfrac{EF}{RQ} = \dfrac{DF}{PQ} = \dfrac{PR}{DE}$ $\;\;(\Delta PRQ \;|||\; \Delta DEF)$

$\dfrac{EF}{PQ} = \dfrac{DF}{RQ} = \dfrac{DE}{PR}$ $\;\;(\Delta PRQ \;|||\; \Delta DEF)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the angles in two triangles are equal, then their sides will be in proportion. Which of the options shows the correct ratios?

STEP: Match up the sides
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta PRQ$ is similar to $\Delta DEF$ , because the matching angles are equal.

If the matching angles in two triangles are equal, then their sides are in proportion. So, when we divide each side of one triangle by its partner in the other triangle, we always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

Since $\Delta {\color{blue}{P}}{\color{blue}{R}}Q \;|||\; \Delta {\color{blue}{D}}{\color{blue}{E}}F$ , we can see that ${\color{blue}{PR}}$ matches with ${\color{blue}{DE}}$ .

In the same way, we can see that ${\color{ForestGreen}{PQ}}$ matches with ${\color{ForestGreen}{DF}}$ $(\Delta {\color{ForestGreen}{P}}R{\color{ForestGreen}{Q}} \;|||\; \Delta {\color{ForestGreen}{D}}E{\color{ForestGreen}{F}})$ .

Finally, the last sides must also match up. So ${\color{red}{RQ}}$ matches with ${\color{red}{EF}}$ $(\Delta P{\color{red}{R}}{\color{red}{Q}} \;|||\; \Delta D{\color{red}{E}}{\color{red}{F}})$ .

TIP: We must always divide the triangle sides in the same order. So, if we put $\Delta DEF$ 's sides in the numerator for the first fraction, then $\Delta DEF$ 's sides must always be in the numerator.

We can also see the matching sides in the diagram:

So, the correct answer is

$\begin{matrix} \frac{E F}{R Q} = \frac{D F}{P Q} = \frac{D E}{P R} (Δ P R Q | | | Δ D E F) \end{matrix}$ $\begin{array}{c} \dfrac{EF}{RQ} = \dfrac{DF}{PQ} = \dfrac{DE}{PR} \;\; (\Delta PRQ \;|||\; \Delta DEF) \end{array}$

NOTE: We only know that the sides are in proportion because the triangles are similar. So, we write this as the reason.

There are lots of correct ways to write the ratios. These are two more correct examples:
1. $\dfrac{RQ}{EF} = \dfrac{PQ}{DF} = \dfrac{PR}{DE} \;\; (\Delta PRQ \;|||\; \Delta DEF)$
2. $\dfrac{DF}{PQ} = \dfrac{DE}{PR} = \dfrac{EF}{RQ} \;\; (\Delta PRQ \;|||\; \Delta DEF)$
Submit your answer as:
$\dfrac{EF}{RQ} = \dfrac{DF}{PQ} = \dfrac{DE}{PR}$ $\;\;(\Delta PRQ \;|||\; \Delta DEF)$

Basic deductions from similar triangles

In the diagram below, $\Delta MNL \;|||\; \Delta ABC$ , because their sides are in proportion.

What can you work out about the angles in the two triangles?
Answer:
- $\angle N$ $\; = \angle$ $\;\;(\Delta MNL \;|||\; \Delta ABC)$
- $\angle L$ $\; = \angle$ $\;\;(\Delta MNL \;|||\; \Delta ABC)$
- $\angle M$ $\; = \angle$ $\;\;(\Delta MNL \;|||\; \Delta ABC)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the sides of two triangles are in proportion, then their matching angles are equal. Can you match up the angles of the two triangles?

STEP: Match up the angles
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta MNL$ is similar to $\Delta ABC$ , because the sides are in proportion. We can see that the all of the sides in $\Delta MNL$ have been multiplied by $\text{2}$ to get the sides in $\Delta ABC$ .

If the sides of two triangles are in proportion, then all of their matching angles will be equal. We can identify the matching angles using the similarity statement: $\Delta MNL \;|||\; \Delta ABC$ .

So,
- ${\color{blue}{\hat{M}}} = {\color{blue}{\hat{A}}} \;\;(\Delta {\color{blue}{M}}NL \;|||\; \Delta {\color{blue}{A}}BC)$
- ${\color{red}{\hat{N}}} = {\color{red}{\hat{B}}} \;\;(\Delta M{\color{red}{N}}L \;|||\; \Delta A{\color{red}{B}}C)$
- ${\color{ForestGreen}{\hat{L}}} = {\color{ForestGreen}{\hat{C}}} \;\;(\Delta MN{\color{ForestGreen}{L}} \;|||\; \Delta AB{\color{ForestGreen}{C}})$
NOTE: We only know that the angles are equal because the triangles are similar. So, we write this as the reason next to the deduction.

Submit your answer as: andand
In the diagram below, $\Delta DFE \;|||\; \Delta PQR$ , because their angles are equal.

What can you deduce about the sides of the two triangles? Choose the correct option.
Answer:
$\dfrac{PQ}{DF} = \dfrac{PR}{DE} = \dfrac{QR}{FE}$ $\;\;(\Delta DFE \;|||\; \Delta PQR)$

$\dfrac{\hat{P}}{\hat{D}} = \dfrac{\hat{Q}}{\hat{F}} = \dfrac{\hat{R}}{\hat{E}}$ $\;\;(\Delta DFE \;|||\; \Delta PQR)$

$\dfrac{PQ}{DF} = \dfrac{PR}{FE} = \dfrac{QR}{DE}$ $\;\;(\Delta DFE \;|||\; \Delta PQR)$

$\dfrac{DF}{PQ} = \dfrac{PR}{DE} = \dfrac{QR}{FE}$ $\;\;(\Delta DFE \;|||\; \Delta PQR)$
HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

If the angles in two triangles are equal, then their sides will be in proportion. Which of the options shows the correct ratios?

STEP: Match up the sides
[−3 points ⇒ 0 / 3 points left]

We have been told that $\Delta DFE$ is similar to $\Delta PQR$ , because the matching angles are equal.

If the matching angles in two triangles are equal, then their sides are in proportion. So, when we divide each side of one triangle by its partner in the other triangle, we always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

Since $\Delta {\color{blue}{D}}{\color{blue}{F}}E \;|||\; \Delta {\color{blue}{P}}{\color{blue}{Q}}R$ , we can see that ${\color{blue}{DF}}$ matches with ${\color{blue}{PQ}}$ .

In the same way, we can see that ${\color{ForestGreen}{DE}}$ matches with ${\color{ForestGreen}{PR}}$ $(\Delta {\color{ForestGreen}{D}}F{\color{ForestGreen}{E}} \;|||\; \Delta {\color{ForestGreen}{P}}Q{\color{ForestGreen}{R}})$ .

Finally, the last sides must also match up. So ${\color{red}{FE}}$ matches with ${\color{red}{QR}}$ $(\Delta D{\color{red}{F}}{\color{red}{E}} \;|||\; \Delta P{\color{red}{Q}}{\color{red}{R}})$ .

TIP: We must always divide the triangle sides in the same order. So, if we put $\Delta PQR$ 's sides in the numerator for the first fraction, then $\Delta PQR$ 's sides must always be in the numerator.

We can also see the matching sides in the diagram:

So, the correct answer is

$\begin{matrix} \frac{P Q}{D F} = \frac{P R}{D E} = \frac{Q R}{F E} (Δ D F E | | | Δ P Q R) \end{matrix}$ $\begin{array}{c} \dfrac{PQ}{DF} = \dfrac{PR}{DE} = \dfrac{QR}{FE} \;\; (\Delta DFE \;|||\; \Delta PQR) \end{array}$

NOTE: We only know that the sides are in proportion because the triangles are similar. So, we write this as the reason.

There are lots of correct ways to write the ratios. These are two more correct examples:
1. $\dfrac{DF}{PQ} = \dfrac{DE}{PR} = \dfrac{FE}{QR} \;\; (\Delta DFE \;|||\; \Delta PQR)$
2. $\dfrac{PR}{DE} = \dfrac{QR}{FE} = \dfrac{PQ}{DF} \;\; (\Delta DFE \;|||\; \Delta PQR)$
Submit your answer as:
$\dfrac{PQ}{DF} = \dfrac{PR}{DE} = \dfrac{QR}{FE}$ $\;\;(\Delta DFE \;|||\; \Delta PQR)$

Identify similar triangles

In the diagram below, $\hat{B} = 36°$ , and $B\hat{C}D = B\hat{E}A = 98°$ .

Identify one triangle that is similar to $\Delta EAB$ .

Answer:

Δ E A B | | | Δ

$\Delta EAB \;|||\; \Delta$

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which other triangle in the diagram has the same angles as $\Delta EAB$ ?

STEP: Identify the similar triangle
[−2 points ⇒ 0 / 2 points left]

We can see that $\hat{B}$ is common to $\Delta EAB$ and $\Delta CDB$ . We were also told that $B\hat{C}D = B\hat{E}A$ .

So, we have two pairs of matching angles in $\Delta EAB$ and in $\Delta CDB$ .

We can use sum of angles in both triangles to calculate that $\hat{A} = \hat{D} = 46°$ .

So all three pairs of matching angles are equal, and $\Delta {\color{ForestGreen}{E}}{\color{blue}{A}}{\color{red}{B}} \;|||\; \Delta {\color{ForestGreen}{C}}{\color{blue}{D}}{\color{red}{B}}$ .

TIP: Whenever we know that two pairs of angles are the same in two triangles, the third pair will automatically also be the same. This is because of the sum of the angles in the two triangles.

Submit your answer as:

Identify similar triangles

In the diagram below, $LM \bot MK$ and $MJ \bot LK$ . Also, $\hat{L} = 36°$ and $\hat{K} = 54°$ .

Identify one triangle that is similar to $\Delta LMK$ .

Answer:

Δ L M K | | | Δ

$\Delta LMK \;|||\; \Delta$

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which other triangle in the diagram has the same angles as $\Delta LMK$ ?

STEP: Identify the similar triangle
[−2 points ⇒ 0 / 2 points left]

There are three triangles in the diagram: $\Delta LMK$ , $\Delta MJK$ , and $\Delta LJM$ . They are all similar to each other.

Looking at $\Delta LMK$ and $\Delta MJK$ , we can see that:

$\hat{K}$ is a common angle.
$L\hat{M}K = K\hat{J}M = 90°$ , which was given.
$K\hat{M}J = \hat{L} = 36°$ , using sum of angles in $\Delta MJK$ .

So, $\Delta {\color{ForestGreen}{L}}{\color{red}{M}}{\color{blue}{K}} \;|||\; \Delta {\color{ForestGreen}{M}}{\color{red}{J}}{\color{blue}{K}}$ , because the angles in both triangles are equal.

We could also prove that $\Delta {\color{ForestGreen}{L}}{\color{red}{M}}{\color{blue}{K}} \;|||\; \Delta {\color{ForestGreen}{L}}{\color{red}{J}}{\color{blue}{M}}$ .

Since $\Delta MJK$ and $\Delta LJM$ are both similar to $\Delta LMK$ , they are also similar to each other.

So, either $\Delta MJK$ or $\Delta LJM$ is a correct answer.

TIP: Whenever we know that two pairs of angles are the same in two triangles, the third pair will automatically also be the same. This is because of the sum of the angles in the two triangles.

Submit your answer as:

Identify similar triangles

In the diagram below, $\hat{R} = 45°$ , and $\hat{S} = \hat{U} = 43°$ .

Identify one triangle that is similar to $\Delta RSP$ .

Answer:

Δ R S P | | | Δ

$\Delta RSP \;|||\; \Delta$

one-of

type(string.nocase)

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Which other triangle in the diagram has the same angles as $\Delta RSP$ ?

STEP: Identify the similar triangle
[−2 points ⇒ 0 / 2 points left]

We can see that $\hat{R}$ is common to $\Delta RUT$ and $\Delta RSP$ . We were also told that $\hat{S} = \hat{U}$ .

So, we have two pairs of matching angles in $\Delta RUT$ and in $\Delta RSP$ .

We can use sum of angles in both triangles to calculate that $R\hat{T}U = R\hat{P}S = 92°$ .

So all three pairs of matching angles are equal, and $\Delta {\color{red}{R}}{\color{ForestGreen}{U}}{\color{blue}{T}} \;|||\; \Delta {\color{red}{R}}{\color{ForestGreen}{S}}{\color{blue}{P}}$ .

TIP: Whenever we know that two pairs of angles are the same in two triangles, the third pair will automatically also be the same. This is because of the sum of the angles in the two triangles.

Submit your answer as:

Prove similarity with parallel lines

In the diagram below, $AD \parallel EC$ .

Prove that $\Delta ADB \;|||\; \Delta ECB$ .

Answer:

In $\Delta ADB$ and $\Delta ECB$ :

$\angle BDA = \angle$ (corresp $\angle$ s; $AD \parallel EC$ )
$\angle BAD = \angle$ (corresp $\angle$ s; $AD \parallel EC$ )

$\therefore \Delta ADB \;|||\; \Delta$ .

string

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if all of their matching angles are equal.

STEP: Prove that the triangles are equiangular
[−3 points ⇒ 1 / 4 points left]

There are two ways to prove that triangles are similar. We can prove that

their matching sides are in proportion (sides of $\Delta$ in prop), or
their matching angles are equal (equiangular $\Delta$ s).

We have not been given any information about sides, so we will not be able to prove that the sides of these triangles are in proportion. But, we can use our geometry reasons to prove that their matching angles are equal.

TIP: When you are told that two lines are parallel, you should look out for alternate, corresponding, and co-interior angles.

In $\Delta ADB$ and $\Delta ECB$ :

${\color{blue}{\angle BDA}} = {\color{blue}{\angle BCE}}$ (corresp $\angle$ s; $AD \parallel EC$ )
${\color{ForestGreen}{\angle BAD}} = {\color{ForestGreen}{\angle BEC}}$ (corresp $\angle$ s; $AD \parallel EC$ )
${\color{red}{\angle B}}$ is common

STEP: Complete the proof by labelling the triangles and giving a reason
[−1 point ⇒ 0 / 4 points left]

We must always match up the triangle vertices on the last line of the congruency proof. This is easier to see when the triangles are drawn separately.

We can see that the angle at $A$ is equal to the angle at $E$ . So, Point ${\color{ForestGreen}{A}}$ matches with Point ${\color{ForestGreen}{E}}$ .

In the same way:

Point ${\color{blue}{D}}$ matches with ${\color{blue}{C}}$ , and
Point ${\color{red}{B}}$ matches with ${\color{red}{B}}$ .

So $\Delta {\color{ForestGreen}{{A}}}{\color{blue}{{D}}}{\color{red}{{B}}} \;|||\; \Delta {\color{ForestGreen}{{E}}}{\color{blue}{{C}}}{\color{red}{{B}}}$ (equiangular $\Delta$ s).

NOTE: You must use the exact reason equiangular $\Delta$ s. This means that each angle in one triangle is equal to an angle in the other triangle.

Submit your answer as: andandandand

Prove similarity with parallel lines

In the diagram below, $CA \parallel EB$ . Also, $CB$ and $AE$ are straight lines.

Prove that $\Delta CAD \;|||\; \Delta BED$ .

Answer:

In $\Delta CAD$ and $\Delta BED$ :

$\angle CDA = \angle BDE$
$\angle A = \angle E$
$\angle C = \angle B$

$\therefore \Delta CAD \;|||\; \Delta$ .

string

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if all of their matching angles are equal.

STEP: Prove that the triangles are equiangular
[−3 points ⇒ 1 / 4 points left]

There are two ways to prove that triangles are similar. We can prove that

their matching sides are in proportion (sides of $\Delta$ in prop), or
their matching angles are equal (equiangular $\Delta$ s).

We have not been given any information about sides, so we will not be able to prove that the sides of these triangles are in proportion. But, we can use our geometry reasons to prove that their matching angles are equal.

TIP: When you are told that two lines are parallel, you should look out for alternate, corresponding, and co-interior angles.

In $\Delta CAD$ and $\Delta BED$ :

${\color{red}{\angle CDA}} = {\color{red}{\angle BDE}}$ (vert opp $\angle$ s equal)
${\color{blue}{\angle A}} = {\color{blue}{\angle E}}$ (alt $\angle$ s; $CA \parallel EB$ )
${\color{ForestGreen}{\angle C}} = {\color{ForestGreen}{\angle B}}$ (alt $\angle$ s; $CA \parallel EB$ )

STEP: Complete the proof by labelling the triangles and giving a reason
[−1 point ⇒ 0 / 4 points left]

We must always match up the triangle vertices on the last line of the congruency proof. This is easier to see when the triangles are drawn separately.

We can see that the angle at $C$ is equal to the angle at $B$ . So, Point ${\color{ForestGreen}{C}}$ matches with Point ${\color{ForestGreen}{B}}$ .

In the same way:

Point ${\color{blue}{A}}$ matches with ${\color{blue}{E}}$ , and
Point ${\color{red}{D}}$ matches with ${\color{red}{D}}$ .

So $\Delta {\color{ForestGreen}{{C}}}{\color{blue}{{A}}}{\color{red}{{D}}} \;|||\; \Delta {\color{ForestGreen}{{B}}}{\color{blue}{{E}}}{\color{red}{{D}}}$ (equiangular $\Delta$ s).

NOTE: You must use the exact reason equiangular $\Delta$ s. This means that each angle in one triangle is equal to an angle in the other triangle.

Submit your answer as: andandandand

Prove similarity with parallel lines

In the diagram below, $SQ \parallel TR$ . Also, $SR$ and $QT$ are straight lines.

Prove that $\Delta SQP \;|||\; \Delta RTP$ .

Answer:

In $\Delta SQP$ and $\Delta RTP$ :

$\angle SPQ = \angle RPT$
$\angle Q = \angle T$
$\angle S = \angle R$

$\therefore \Delta SQP \;|||\; \Delta$ .

string

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if all of their matching angles are equal.

STEP: Prove that the triangles are equiangular
[−3 points ⇒ 1 / 4 points left]

There are two ways to prove that triangles are similar. We can prove that

their matching sides are in proportion (sides of $\Delta$ in prop), or
their matching angles are equal (equiangular $\Delta$ s).

We have not been given any information about sides, so we will not be able to prove that the sides of these triangles are in proportion. But, we can use our geometry reasons to prove that their matching angles are equal.

TIP: When you are told that two lines are parallel, you should look out for alternate, corresponding, and co-interior angles.

In $\Delta SQP$ and $\Delta RTP$ :

${\color{red}{\angle SPQ}} = {\color{red}{\angle RPT}}$ (vert opp $\angle$ s equal)
${\color{blue}{\angle Q}} = {\color{blue}{\angle T}}$ (alt $\angle$ s; $SQ \parallel TR$ )
${\color{ForestGreen}{\angle S}} = {\color{ForestGreen}{\angle R}}$ (alt $\angle$ s; $SQ \parallel TR$ )

STEP: Complete the proof by labelling the triangles and giving a reason
[−1 point ⇒ 0 / 4 points left]

We must always match up the triangle vertices on the last line of the congruency proof. This is easier to see when the triangles are drawn separately.

We can see that the angle at $S$ is equal to the angle at $R$ . So, Point ${\color{ForestGreen}{S}}$ matches with Point ${\color{ForestGreen}{R}}$ .

In the same way:

Point ${\color{blue}{Q}}$ matches with ${\color{blue}{T}}$ , and
Point ${\color{red}{P}}$ matches with ${\color{red}{P}}$ .

So $\Delta {\color{ForestGreen}{{S}}}{\color{blue}{{Q}}}{\color{red}{{P}}} \;|||\; \Delta {\color{ForestGreen}{{R}}}{\color{blue}{{T}}}{\color{red}{{P}}}$ (equiangular $\Delta$ s).

NOTE: You must use the exact reason equiangular $\Delta$ s. This means that each angle in one triangle is equal to an angle in the other triangle.

Submit your answer as: andandandand

Similar triangles

Given the following diagrams:

Diagram A

Diagram B

Which diagram shows a pair of similar triangles?

Answer: The similar triangles are in Diagram .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

In maths, the word 'similar' means that two objects have the same shape. Look at the triangles shown and try to determine which pair of triangles are the same shape as each other.

STEP: Compare the triangles and decide which ones are similar
[−2 points ⇒ 0 / 2 points left]

There are two ways we can check for similarity in triangles: using the sides or using the angles.

Angles (AAA): If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Sides (SSS): If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar.

We note that we are given information about the sides of all the triangles and so we need to find a diagram that shows a pair of triangles with all three pairs of corresponding sides in proportion.

We look at the side labels. In Diagram A we note the three pairs of corresponding sides are in proportion. In Diagram B we note that the three pairs of corresponding sides are in different proportions.

Therefore Diagram A gives a pair of triangles that are similar. When two shapes are similar, we use the symbol $\; ||| \;$ to show it. For example, in Diagram A, $\Delta ACB ||| \Delta DFE$ , which means, 'triangle $ACB$ is similar to triangle $DFE$ .'

The correct choice is A.

Submit your answer as:

Similar triangles

Given the following diagrams:

Diagram A

Diagram B

Which diagram shows a pair of similar triangles?

Answer: The similar triangles are in Diagram .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

In maths, the word 'similar' means that two objects have the same shape. Look at the triangles shown and try to determine which pair of triangles are the same shape as each other.

STEP: Compare the triangles and decide which ones are similar
[−2 points ⇒ 0 / 2 points left]

There are two ways we can check for similarity in triangles: using the sides or using the angles.

Angles (AAA): If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Sides (SSS): If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar.

We note that we are given information about the sides of all the triangles and so we need to find a diagram that shows a pair of triangles with all three pairs of corresponding sides in proportion.

We look at the side labels. In Diagram A we note the three pairs of corresponding sides are in proportion. In Diagram B we note that the three pairs of corresponding sides are in different proportions.

Therefore Diagram A gives a pair of triangles that are similar. When two shapes are similar, we use the symbol $\; ||| \;$ to show it. For example, in Diagram A, $\Delta ACB ||| \Delta DFE$ , which means, 'triangle $ACB$ is similar to triangle $DFE$ .'

The correct choice is A.

Submit your answer as:

Similar triangles

Given the following diagrams:

Diagram A

Diagram B

Which diagram shows a pair of similar triangles?

Answer: The similar triangles are in Diagram .

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

In maths, the word 'similar' means that two objects have the same shape. Look at the triangles shown and try to determine which pair of triangles are the same shape as each other.

STEP: Compare the triangles and decide which ones are similar
[−2 points ⇒ 0 / 2 points left]

There are two ways we can check for similarity in triangles: using the sides or using the angles.

Angles (AAA): If all three pairs of corresponding angles of two triangles are equal, then the triangles are similar.
Sides (SSS): If all three pairs of corresponding sides of two triangles are in proportion, then the triangles are similar.

We note that we are given information about the sides of all the triangles and so we need to find a diagram that shows a pair of triangles with all three pairs of corresponding sides in proportion.

We look at the side labels. In Diagram A we note the three pairs of corresponding sides are in proportion. In Diagram B we note that the three pairs of corresponding sides are in different proportions.

Therefore Diagram A gives a pair of triangles that are similar. When two shapes are similar, we use the symbol $\; ||| \;$ to show it. For example, in Diagram A, $\Delta ACB ||| \Delta DFE$ , which means, 'triangle $ACB$ is similar to triangle $DFE$ .'

The correct choice is A.

Submit your answer as:

Prove simple similarity

Consider the triangles given below:

Select the correct options to prove that $\Delta YZX \;|||\; \Delta FED$ .

Answer:

In $\Delta YZX$ and $\Delta FED$ :

	Step	Reason
1.	$\angle Z = \angle E = 81°$	(given)
2.	$\angle X = 52°$	(given)
	$= 52°$
	$\therefore$
3.	$\angle F = 47°$	(given)
	$= 47°$
	$\therefore$
	$\therefore \Delta YZX \;\|\|\|\; \Delta FED$

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

There are two ways of proving similarity:

Prove that the two triangles are equiangular.
Prove that the two triangles have sides in proportion.

Which would be more relevant here?

STEP: Find missing angles
[−4 points ⇒ 1 / 5 points left]

In these two triangles, we have more information about the angles than about the sides. So, we will prove that the two triangles are equiangular.

In order to do this, first we need to find the missing angles.

In $\Delta ZXY$ :

$\hat{Y} = 180° - 81° - 52°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{Y} = 47°$

In $\Delta EDF$ :

$\hat{D} = 180° - 81° - 47°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{D} = 52°$

STEP: Complete the proof
[−1 point ⇒ 0 / 5 points left]

Since we have found three pairs of equal angles, we know that the two triangles are similar. Now we need to write our conclusion as a formal proof:

In $\Delta YZX$ and $\Delta FED$ :

${\color{red}{\hat{Z}}} = {\color{red}{\hat{E}}} = 81°$ (given)
$\hat{X} = 52°$ (given)
$\quad$ $\hat{D} = 52°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{ForestGreen}{\hat{X}}} = {\color{ForestGreen}{\hat{D}}}$
$\hat{F} = 47°$ (given)
$\quad$ $\hat{Y} = 47°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{blue}{\hat{F}}} = {\color{blue}{\hat{Y}}}$

$\therefore \Delta YZX \;|||\; \Delta FED$ (equiangular $\Delta$ s)

Submit your answer as: andandandandandand

Prove simple similarity

Consider the triangles given below. $ZV$ and $YW$ are straight lines.

Select the correct options to prove that $\Delta ZXW \;|||\; \Delta YXV$ .

Answer:

In $\Delta ZXW$ and $\Delta YXV$ :

	Step	Reason
1.	$\angle Z = \angle Y = 72°$	(given)
2.	$\angle YXV = 41°$	(given)
	$= 41°$
	$\therefore$
3.	$\angle V = 67°$	(sum of $\angle$ s in $\Delta$ )
	$= 67°$
	$\therefore$
	$\therefore \Delta ZXW \;\|\|\|\; \Delta YXV$

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

There are two ways of proving similarity:

Prove that the two triangles are equiangular.
Prove that the two triangles have sides in proportion.

Which would be more relevant here?

STEP: Find missing angles
[−4 points ⇒ 1 / 5 points left]

In these two triangles, we have more information about the angles than about the sides. So, we will prove that the two triangles are equiangular.

In order to do this, first we need to find the missing angles.

At Point $X$ , two straight lines intersect with each other.

$\therefore Y\hat{X}V = Z\hat{X}W = 41°$ $(\text{vert opp } \angle \text{s equal})$

In $\Delta YXV$ :

$\hat{V} = 180° - 72° - 41°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{V} = 67°$

In $\Delta ZXW$ :

$\hat{W} = 180° - 72° - 41°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{W} = 67°$

STEP: Complete the proof
[−1 point ⇒ 0 / 5 points left]

Since we have found three pairs of equal angles, we know that the two triangles are similar. Now we need to write our conclusion as a formal proof:

In $\Delta ZXW$ and $\Delta YXV$ :

${\color{ForestGreen}{\hat{Z}}} = {\color{ForestGreen}{\hat{Y}}} = 72°$ (given)
$Y\hat{X}V = 41°$ (given)
$\quad$ $Z\hat{X}W = 41°$ (vert opp $\angle$ s equal)
$\quad$ $\therefore {\color{blue}{Y\hat{X}V}} = {\color{blue}{Z\hat{X}W}}$
$\hat{V} = 67°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\hat{W} = 67°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{red}{\hat{V}}} = {\color{red}{\hat{W}}}$

$\therefore \Delta ZXW \;|||\; \Delta YXV$ (equiangular $\Delta$ s)

Submit your answer as: andandandandandand

Prove simple similarity

Consider the triangles given below. $SR$ and $TP$ are straight lines.

Select the correct options to prove that $\Delta SQP \;|||\; \Delta TQR$ .

Answer:

In $\Delta SQP$ and $\Delta TQR$ :

	Step	Reason
1.	$\angle S = \angle T = 52°$	(given)
2.	$\angle TQR = 84°$	(given)
	$= 84°$
	$\therefore$
3.	$\angle R = 44°$	(sum of $\angle$ s in $\Delta$ )
	$= 44°$
	$\therefore$
	$\therefore \Delta SQP \;\|\|\|\; \Delta TQR$

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

There are two ways of proving similarity:

Prove that the two triangles are equiangular.
Prove that the two triangles have sides in proportion.

Which would be more relevant here?

STEP: Find missing angles
[−4 points ⇒ 1 / 5 points left]

In these two triangles, we have more information about the angles than about the sides. So, we will prove that the two triangles are equiangular.

In order to do this, first we need to find the missing angles.

At Point $Q$ , two straight lines intersect with each other.

$\therefore T\hat{Q}R = S\hat{Q}P = 84°$ $(\text{vert opp } \angle \text{s equal})$

In $\Delta TQR$ :

$\hat{R} = 180° - 52° - 84°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{R} = 44°$

In $\Delta SQP$ :

$\hat{P} = 180° - 52° - 84°$ $(\text{sum of } \angle \text{s in } \Delta)$
$\therefore \hat{P} = 44°$

STEP: Complete the proof
[−1 point ⇒ 0 / 5 points left]

Since we have found three pairs of equal angles, we know that the two triangles are similar. Now we need to write our conclusion as a formal proof:

In $\Delta SQP$ and $\Delta TQR$ :

${\color{ForestGreen}{\hat{S}}} = {\color{ForestGreen}{\hat{T}}} = 52°$ (given)
$T\hat{Q}R = 84°$ (given)
$\quad$ $S\hat{Q}P = 84°$ (vert opp $\angle$ s equal)
$\quad$ $\therefore {\color{blue}{T\hat{Q}R}} = {\color{blue}{S\hat{Q}P}}$
$\hat{R} = 44°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\hat{P} = 44°$ (sum of $\angle$ s in $\Delta$ )
$\quad$ $\therefore {\color{red}{\hat{R}}} = {\color{red}{\hat{P}}}$

$\therefore \Delta SQP \;|||\; \Delta TQR$ (equiangular $\Delta$ s)

Submit your answer as: andandandandandand

Identifying similar and congruent polygons

Polygon $P$ is drawn below:

Which of the following is similar to Polygon $P$ ?

Answer:

A
B
C
D

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other.

STEP: Choose the similar polygon
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other. For two polygons to be similar, they must have:

the same number of sides,
all of the matching sides in proportion, and
all of the matching angles equal in size.

NOTE: In normal English, we say two things are similar if they are almost the same. But, in Maths, it is not precise enough to say that similar shapes are approximately the same. In Maths, similarity has a very specific meaning.

Option A is the same shape as the given polygon, even though it is a different size.

So, Option A is similar to the given polygon.

Submit your answer as:

A

Identifying similar and congruent polygons

Polygon $P$ is drawn below:

Which of the following is similar to Polygon $P$ ?

Answer:

A
B
C
D

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other.

STEP: Choose the similar polygon
[−2 points ⇒ 0 / 2 points left]

Two polygons are similar if they are exactly the same shape, but one might be bigger or smaller than the other. For two polygons to be similar, they must have:

the same number of sides,
all of the matching sides in proportion, and
all of the matching angles equal in size.

NOTE: In normal English, we say two things are similar if they are almost the same. But, in Maths, it is not precise enough to say that similar shapes are approximately the same. In Maths, similarity has a very specific meaning.

Option B is the same shape as the given polygon, even though it is a different size.

So, Option B is similar to the given polygon.

Submit your answer as:

B

Identifying similar and congruent polygons

Polygon $P$ is drawn below:

Which of the following is congruent to Polygon $P$ ?

Answer:

A
B
C
D

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Two polygons are congruent if they are exactly the same size and shape. If we cut out two congruent polygons, one could be placed exactly on top of the other.

STEP: Choose the congruent polygon
[−2 points ⇒ 0 / 2 points left]

Two polygons are congruent if they are exactly the same size and shape. For two polygons to be congruent, they must have:

the same number of sides,
all of the matching sides equal in length, and
all of the matching angles equal in size.

NOTE: If two polygons are the same shape, but different sizes, then they are similar. But they are not congruent. We could not place one polygon exactly on top of the other!

Option D is exactly the same size and shape as the given polygon. This is true even though they are in different orientations.

TIP: Congruent (or similar) polygons can have a different orientation. This just means that they are facing different directions. If we cut them out and turn them we could still place one directly on top of the other.

Submit your answer as:

D

Calculate a side in similar polygons

The two polygons drawn below are similar.

Determine the value of $m$ .

Answer:

m =

$m=$

units

$\text{units}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon.

STEP: Determine the scale factor
[−2 points ⇒ 0 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon. This number is called the scale factor.

The sides in the first polygon have been multiplied by $\dfrac{\text{1}}{\text{2}}$ to give the matching sides in the second polygon. For example:

${\color{ForestGreen}{\text{10}}} \times \dfrac{\text{1}}{\text{2}} = {\color{ForestGreen}{\text{5}}}$
${\color{MediumVioletRed}{\text{12}}} \times \dfrac{\text{1}}{\text{2}} = {\color{MediumVioletRed}{\text{6}}}$

The side labelled $m$ matches with the side equal to ${\color{red}{\text{8}}}$ in the first polygon. So:

\begin{aligned} 8 \times \frac{1}{2} & = m \\ ∴ m & = 4 units \end{aligned}

$\begin{aligned} {\color{red}{\text{8}}} \times \dfrac{\text{1}}{\text{2}} &= {\color{red}{m}} \\ \therefore m &= \text{4}\ \text{units} \end{aligned}$

NOTE: We can see that the sides in the first polygon have all been divided by $\text{2}$ to get the sides in the second polygon. Dividing by

2

$\text{2}$ is the same as multiplying by

\frac{1}{2}

$\dfrac{1}{2}$ .

Submit your answer as:

Calculate a side in similar polygons

The two polygons drawn below are similar.

Determine the value of $a$ .

Answer:

a =

$a=$

units

$\text{units}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon.

STEP: Determine the scale factor
[−2 points ⇒ 0 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon. This number is called the scale factor.

The sides in the first polygon have been multiplied by $\dfrac{\text{1}}{\text{2}}$ to give the matching sides in the second polygon. For example:

${\color{teal}{\text{18}}} \times \dfrac{\text{1}}{\text{2}} = {\color{teal}{\text{9}}}$
${\color{DarkGray}{\text{12}}} \times \dfrac{\text{1}}{\text{2}} = {\color{DarkGray}{\text{6}}}$

The side labelled $a$ matches with the side equal to ${\color{blue}{\text{16}}}$ in the first polygon. So:

\begin{aligned} 16 \times \frac{1}{2} & = a \\ ∴ a & = 8 units \end{aligned}

$\begin{aligned} {\color{blue}{\text{16}}} \times \dfrac{\text{1}}{\text{2}} &= {\color{blue}{a}} \\ \therefore a &= \text{8}\ \text{units} \end{aligned}$

NOTE: We can see that the sides in the first polygon have all been divided by $\text{2}$ to get the sides in the second polygon. Dividing by

2

$\text{2}$ is the same as multiplying by

\frac{1}{2}

$\dfrac{1}{2}$ .

Submit your answer as:

Calculate a side in similar polygons

The two polygons drawn below are similar.

Determine the value of $n$ .

Answer:

n =

$n=$

units

$\text{units}$

numeric

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon.

STEP: Determine the scale factor
[−2 points ⇒ 0 / 2 points left]

If two polygons are similar, then their sides are in proportion. This means that every side in the first polygon has been multiplied by the same number to get the matching sides in the second polygon. This number is called the scale factor.

The sides in the first polygon have been multiplied by $\text{2}$ to give the matching sides in the second polygon. For example:

${\color{MediumVioletRed}{\text{10}}} \times \text{2} = {\color{MediumVioletRed}{\text{20}}}$
${\color{red}{\text{4}}} \times \text{2} = {\color{red}{\text{8}}}$

The side labelled $n$ matches with the side equal to ${\color{blue}{\text{5}}}$ in the first polygon. So:

\begin{aligned} 5 \times 2 & = n \\ ∴ n & = 10 units \end{aligned}

$\begin{aligned} {\color{blue}{\text{5}}} \times \text{2} &= {\color{blue}{n}} \\ \therefore n &= \text{10}\ \text{units} \end{aligned}$

Submit your answer as:

Similarity or congruency?

Consider the diagram below and answer the questions that follow:

Erioluwa needs to prove that $\Delta YXZ \;|||\; \Delta ABC$ . She has already started, and her incomplete proof is written below:

In $\Delta YXZ$ and $\Delta ABC$ :
1. $\hat{Y} = \hat{A}$ (given)
  ...
Answer:
How should Erioluwa complete her proof? Choose the best option.
Option A
2. $\hat{Z} = 51°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{Z} = \hat{C}$
3. Also, $\hat{B} = 40°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{X} = \hat{B}$
$\therefore \Delta YXZ \;|||\; \Delta ABC$ (equiangular $\Delta$ s)

Option B
2. $\hat{X} = \hat{B}$ (given)
3. $\hat{Z} = 51°$ (sum of $\angle$ s in $\Delta$ )
Also, $\hat{C} = 51°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{Z} = \hat{C}$
$\therefore \Delta YXZ \;|||\; \Delta ABC$ (equiangular $\Delta$ s)

Option C
2. $\hat{B} = 40°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{X} = \hat{B}$
3. $\dfrac{YX}{AC} = \dfrac{12}{12} = 1$
$\therefore \Delta YXZ \;|||\; \Delta ABC$ (sides of $\Delta$ in prop)

Option D
2. $\hat{Z} = 51°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{Z} = \hat{C}$
3. $YX = CA$ (given)
$\therefore \Delta YXZ \;|||\; \Delta ABC$ (SAA)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if they are the same shape, but they may be a different size. Their angles will be equal and their sides will be in proportion.

Two triangles are congruent if they are exactly the same size and shape. Their matching sides and angles will all be equal.

STEP: Choose the correct proof
[−4 points ⇒ 0 / 4 points left]

✔ Option A

This option uses sum of $\angle$ s in $\Delta$ correctly to prove that there are three pairs of matching angles in the triangles.

✘ Option B

This "proof" makes mistakes on Step 2 and Step 3. We were not given that $\hat{X} = \hat{B}$ ; we must prove this using sum of $\angle$ s in $\Delta$ . Also, on Step 3, we know that $\hat{C} = 51°$ because this was given , not because of sum of $\angle$ s in $\Delta$ .

✘ Option C

To prove that two triangles are similar, you must either prove that all of the sides are in proportion, or that all the matching angles are equal . This proof tries to do a mixture of these two, which is not allowed.

✘ Option D

We cannot prove that the triangles are congruent, even though we can prove a pair of equal sides and two pairs of equal angles. This is because the angles are not in the same positions relative to the given side. The information we have been given actually proves that the triangles are slightly different sizes.

TIP: The question told us to prove that the triangles are similar (using the symbol $|||$ ). You should always read the question statement very carefully.

The similarity is easier to see when the triangles are in the same orientation:

The full proof is shown below:

In $\color{gray}{\Delta YXZ}$ and $\color{gray}{\Delta ABC}$ :
1. $\color{gray}{\hat{Y} = \hat{A}}$ (given)
2. $\hat{Z} = 51°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{Z} = \hat{C}$
3. Also, $\hat{B} = 40°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{X} = \hat{B}$
$\therefore \Delta YXZ \;|||\; \Delta ABC$ (equiangular $\Delta$ s)

Submit your answer as:
Option A
2. $\hat{Z} = 51°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{Z} = \hat{C}$
3. Also, $\hat{B} = 40°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{X} = \hat{B}$
$\therefore \Delta YXZ \;|||\; \Delta ABC$ (equiangular $\Delta$ s)
Hence determine the length of $AB$ , giving a reason for your answer.

INSTRUCTION: Do not round off your answer.

Answer: $AB =$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $AB$ . So, start by writing $AB$ in the numerator:

$\begin{aligned} \frac{A B}{◻} = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{AB}{\Box} = \dfrac{\Box}{\Box} \end{aligned}$

Since $\Delta {\color{teal}{Y}}{\color{teal}{X}}Z \;|||\; \Delta {\color{teal}{A}}{\color{teal}{B}}C$ , we can see that ${\color{teal}{AB}}$ matches with ${\color{teal}{YX}}$ .

So we have:

$\begin{aligned} \frac{A B}{Y X} = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{AB}{YX} = \dfrac{\Box}{\Box} \end{aligned}$

We were also given information about $CA$ and $YZ$ . Since $\Delta {\color{orange}{Y}}X{\color{orange}{Z}} \;|||\; \Delta {\color{orange}{A}}B{\color{orange}{C}}$ , we can see that ${\color{orange}{CA}}$ matches with ${\color{orange}{YZ}}$ .

So,

$\begin{aligned} \frac{A B}{Y X} = \frac{C A}{Y Z} (Δ Y X Z | | | Δ A B C) \end{aligned}$ $\begin{aligned} \dfrac{AB}{YX} = \dfrac{CA}{YZ} \; (\Delta YXZ \;|||\; \Delta ABC) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So we write this as the reason. Remember, sides of $\Delta$ in prop is used to prove that triangles are similar, not to determine side lengths.

STEP: Solve the equation using inverse operations
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \frac{A B}{12} & = \frac{12}{10} \\ ∴ \frac{A B}{12} \times 12 & = \frac{12}{10} \times 12 \\ ∴ A B & = \frac{72}{5} \\ = 14,4 \end{aligned}$ $\begin{aligned} \dfrac{{\color{teal}{AB}}}{{\color{teal}{12}}} & = \dfrac{{\color{orange}{12}}}{{\color{orange}{10}}} \\ \therefore \dfrac{AB}{12} \times 12 & = \dfrac{12}{10} \times 12 \\ \therefore AB & = \frac{72}{5}\\ & = \text{14,4} \end{aligned}$

Submit your answer as: and

Similarity or congruency?

Consider the diagram below and answer the questions that follow:

Eniola needs to prove that $\Delta MNL \;|||\; \Delta BCA$ . She has already started, and her incomplete proof is written below:

In $\Delta MNL$ and $\Delta BCA$ :
1. $\hat{M} = \hat{B}$ (given)
  ...
Answer:
How should Eniola complete her proof? Choose the best option.
Option A
2. $\hat{N} = \hat{C}$ (given)
3. $\hat{L} = 84°$ (sum of $\angle$ s in $\Delta$ )
Also, $\hat{A} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{L} = \hat{A}$
$\therefore \Delta MNL \;|||\; \Delta BCA$ (equiangular $\Delta$ s)

Option B
2. $\hat{L} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{L} = \hat{A}$
3. Also, $\hat{C} = 41°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{N} = \hat{C}$
$\therefore \Delta MNL \;|||\; \Delta BCA$ (equiangular $\Delta$ s)

Option C
2. $\hat{C} = 41°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{N} = \hat{C}$
3. $\dfrac{MN}{BA} = \dfrac{15}{15} = 1$
$\therefore \Delta MNL \;|||\; \Delta BCA$ (sides of $\Delta$ in prop)

Option D
2. $\hat{L} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{L} = \hat{A}$
3. $MN = AB$ (given)
$\therefore \Delta MNL \;|||\; \Delta BCA$ (SAA)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if they are the same shape, but they may be a different size. Their angles will be equal and their sides will be in proportion.

Two triangles are congruent if they are exactly the same size and shape. Their matching sides and angles will all be equal.

STEP: Choose the correct proof
[−4 points ⇒ 0 / 4 points left]

✘ Option A

This "proof" makes mistakes on Step 2 and Step 3. We were not given that $\hat{N} = \hat{C}$ ; we must prove this using sum of $\angle$ s in $\Delta$ . Also, on Step 3, we know that $\hat{A} = 84°$ because this was given , not because of sum of $\angle$ s in $\Delta$ .

✔ Option B

This option uses sum of $\angle$ s in $\Delta$ correctly to prove that there are three pairs of matching angles in the triangles.

✘ Option C

To prove that two triangles are similar, you must either prove that all of the sides are in proportion, or that all the matching angles are equal . This proof tries to do a mixture of these two, which is not allowed.

✘ Option D

We cannot prove that the triangles are congruent, even though we can prove a pair of equal sides and two pairs of equal angles. This is because the angles are not in the same positions relative to the given side. The information we have been given actually proves that the triangles are slightly different sizes.

TIP: The question told us to prove that the triangles are similar (using the symbol $|||$ ). You should always read the question statement very carefully.

The similarity is easier to see when the triangles are in the same orientation:

The full proof is shown below:

In $\color{gray}{\Delta MNL}$ and $\color{gray}{\Delta BCA}$ :
1. $\color{gray}{\hat{M} = \hat{B}}$ (given)
2. $\hat{L} = 84°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{L} = \hat{A}$
3. Also, $\hat{C} = 41°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{N} = \hat{C}$
$\therefore \Delta MNL \;|||\; \Delta BCA$ (equiangular $\Delta$ s)

Submit your answer as:
Option B
2. $\hat{L} = 84°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{L} = \hat{A}$
3. Also, $\hat{C} = 41°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{N} = \hat{C}$
$\therefore \Delta MNL \;|||\; \Delta BCA$ (equiangular $\Delta$ s)
Hence determine the length of $BC$ , giving a reason for your answer.

INSTRUCTION: Do not round off your answer.

Answer: $BC =$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $BC$ . So, start by writing $BC$ in the numerator:

$\begin{aligned} \frac{B C}{◻} = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{BC}{\Box} = \dfrac{\Box}{\Box} \end{aligned}$

Since $\Delta {\color{teal}{M}}{\color{teal}{N}}L \;|||\; \Delta {\color{teal}{B}}{\color{teal}{C}}A$ , we can see that ${\color{teal}{BC}}$ matches with ${\color{teal}{MN}}$ .

So we have:

$\begin{aligned} \frac{B C}{M N} = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{BC}{MN} = \dfrac{\Box}{\Box} \end{aligned}$

We were also given information about $AB$ and $ML$ . Since $\Delta {\color{orange}{M}}N{\color{orange}{L}} \;|||\; \Delta {\color{orange}{B}}C{\color{orange}{A}}$ , we can see that ${\color{orange}{AB}}$ matches with ${\color{orange}{ML}}$ .

So,

$\begin{aligned} \frac{B C}{M N} = \frac{A B}{M L} (Δ M N L | | | Δ B C A) \end{aligned}$ $\begin{aligned} \dfrac{BC}{MN} = \dfrac{AB}{ML} \; (\Delta MNL \;|||\; \Delta BCA) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So we write this as the reason. Remember, sides of $\Delta$ in prop is used to prove that triangles are similar, not to determine side lengths.

STEP: Solve the equation using inverse operations
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \frac{B C}{15} & = \frac{15}{10} \\ ∴ \frac{B C}{15} \times 15 & = \frac{15}{10} \times 15 \\ ∴ B C & = \frac{45}{2} \\ = 22,5 \end{aligned}$ $\begin{aligned} \dfrac{{\color{teal}{BC}}}{{\color{teal}{15}}} & = \dfrac{{\color{orange}{15}}}{{\color{orange}{10}}} \\ \therefore \dfrac{BC}{15} \times 15 & = \dfrac{15}{10} \times 15 \\ \therefore BC & = \frac{45}{2}\\ & = \text{22,5} \end{aligned}$

Submit your answer as: and

Similarity or congruency?

Consider the diagram below and answer the questions that follow:

Nnenne needs to prove that $\Delta STU \;|||\; \Delta FDE$ . She has already started, and her incomplete proof is written below:

In $\Delta STU$ and $\Delta FDE$ :
1. $\hat{S} = \hat{F}$ (given)
  ...
Answer:
How should Nnenne complete her proof? Choose the best option.
Option A
2. $\hat{U} = 92°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{E}$
3. $ST = EF$ (given)
$\therefore \Delta STU \;|||\; \Delta FDE$ (SAA)

Option B
2. $\hat{D} = 45°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = \hat{D}$
3. $\dfrac{ST}{FE} = \dfrac{14}{14} = 1$
$\therefore \Delta STU \;|||\; \Delta FDE$ (sides of $\Delta$ in prop)

Option C
2. $\hat{U} = 92°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{E}$
3. Also, $\hat{D} = 45°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = \hat{D}$
$\therefore \Delta STU \;|||\; \Delta FDE$ (equiangular $\Delta$ s)

Option D
2. $\hat{T} = \hat{D}$ (given)
3. $\hat{U} = 92°$ (sum of $\angle$ s in $\Delta$ )
Also, $\hat{E} = 92°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{E}$
$\therefore \Delta STU \;|||\; \Delta FDE$ (equiangular $\Delta$ s)
HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

Two triangles are similar if they are the same shape, but they may be a different size. Their angles will be equal and their sides will be in proportion.

Two triangles are congruent if they are exactly the same size and shape. Their matching sides and angles will all be equal.

STEP: Choose the correct proof
[−4 points ⇒ 0 / 4 points left]

✘ Option A

We cannot prove that the triangles are congruent, even though we can prove a pair of equal sides and two pairs of equal angles. This is because the angles are not in the same positions relative to the given side. The information we have been given actually proves that the triangles are slightly different sizes.

✘ Option B

To prove that two triangles are similar, you must either prove that all of the sides are in proportion, or that all the matching angles are equal . This proof tries to do a mixture of these two, which is not allowed.

✔ Option C

This option uses sum of $\angle$ s in $\Delta$ correctly to prove that there are three pairs of matching angles in the triangles.

✘ Option D

This "proof" makes mistakes on Step 2 and Step 3. We were not given that $\hat{T} = \hat{D}$ ; we must prove this using sum of $\angle$ s in $\Delta$ . Also, on Step 3, we know that $\hat{E} = 92°$ because this was given , not because of sum of $\angle$ s in $\Delta$ .

TIP: The question told us to prove that the triangles are similar (using the symbol $|||$ ). You should always read the question statement very carefully.

The similarity is easier to see when the triangles are in the same orientation:

The full proof is shown below:

In $\color{gray}{\Delta STU}$ and $\color{gray}{\Delta FDE}$ :
1. $\color{gray}{\hat{S} = \hat{F}}$ (given)
2. $\hat{U} = 92°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{U} = \hat{E}$
3. Also, $\hat{D} = 45°$ (sum of $\angle$ s in $\Delta$ )
  $\therefore \hat{T} = \hat{D}$
$\therefore \Delta STU \;|||\; \Delta FDE$ (equiangular $\Delta$ s)

Submit your answer as:
Option C
2. $\hat{U} = 92°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{U} = \hat{E}$
3. Also, $\hat{D} = 45°$ (sum of $\angle$ s in $\Delta$ )
$\therefore \hat{T} = \hat{D}$
$\therefore \Delta STU \;|||\; \Delta FDE$ (equiangular $\Delta$ s)
Hence determine the length of $FD$ , giving a reason for your answer.

INSTRUCTION: Do not round off your answer.

Answer: $FD =$
one-of
type(numeric.abserror(0.005))

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Use the similarity statement to choose the correct pair of ratios.

STEP: Use the similarity statement to choose the correct pair of ratios
[−2 points ⇒ 1 / 3 points left]

If two triangles are similar, their sides are in proportion. This means that when you divide the matching sides by each other, you always get the same number.

TIP: Use the similarity statement to choose the correct pair of ratios.

We are looking for $FD$ . So, start by writing $FD$ in the numerator:

$\begin{aligned} \frac{F D}{◻} = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{FD}{\Box} = \dfrac{\Box}{\Box} \end{aligned}$

Since $\Delta {\color{teal}{S}}{\color{teal}{T}}U \;|||\; \Delta {\color{teal}{F}}{\color{teal}{D}}E$ , we can see that ${\color{teal}{FD}}$ matches with ${\color{teal}{ST}}$ .

So we have:

$\begin{aligned} \frac{F D}{S T} = \frac{◻}{◻} \end{aligned}$ $\begin{aligned} \dfrac{FD}{ST} = \dfrac{\Box}{\Box} \end{aligned}$

We were also given information about $EF$ and $SU$ . Since $\Delta {\color{orange}{S}}T{\color{orange}{U}} \;|||\; \Delta {\color{orange}{F}}D{\color{orange}{E}}$ , we can see that ${\color{orange}{EF}}$ matches with ${\color{orange}{SU}}$ .

So,

$\begin{aligned} \frac{F D}{S T} = \frac{E F}{S U} (Δ S T U | | | Δ F D E) \end{aligned}$ $\begin{aligned} \dfrac{FD}{ST} = \dfrac{EF}{SU} \; (\Delta STU \;|||\; \Delta FDE) \end{aligned}$

NOTE: We only know that the fractions are equal because the triangles are similar. So we write this as the reason. Remember, sides of $\Delta$ in prop is used to prove that triangles are similar, not to determine side lengths.

STEP: Solve the equation using inverse operations
[−1 point ⇒ 0 / 3 points left]

$\begin{aligned} \frac{F D}{14} & = \frac{14}{10} \\ ∴ \frac{F D}{14} \times 14 & = \frac{14}{10} \times 14 \\ ∴ F D & = \frac{98}{5} \\ = 19,6 \end{aligned}$ $\begin{aligned} \dfrac{{\color{teal}{FD}}}{{\color{teal}{14}}} & = \dfrac{{\color{orange}{14}}}{{\color{orange}{10}}} \\ \therefore \dfrac{FD}{14} \times 14 & = \dfrac{14}{10} \times 14 \\ \therefore FD & = \frac{98}{5}\\ & = \text{19,6} \end{aligned}$

Submit your answer as: and

Side in $\Delta BAD$	Matching side in $\Delta CAB$
$BA$
$BD$
$AD$

	Step	Reason
1.	$\angle L = \angle F = 55°$	(given)
2.	$\angle J = 65°$	(given)
	$= 65°$
	$\therefore$
3.	$\angle D = 60°$	(given)
	$= 60°$
	$\therefore$
	$\therefore \Delta KLJ \;\|\|\|\; \Delta DFE$

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